Programming FAQ
***************


General Questions
=================


Is there a source code level debugger with breakpoints, single-stepping, etc.?
------------------------------------------------------------------------------

Yes.

Several debuggers for Python are described below, and the built-in
function "breakpoint()" allows you to drop into any of them.

The pdb module is a simple but adequate console-mode debugger for
Python. It is part of the standard Python library, and is "documented
in the Library Reference Manual". You can also write your own debugger
by using the code for pdb as an example.

The IDLE interactive development environment, which is part of the
standard Python distribution (normally available as
Tools/scripts/idle3), includes a graphical debugger.

PythonWin is a Python IDE that includes a GUI debugger based on pdb.
The PythonWin debugger colors breakpoints and has quite a few cool
features such as debugging non-PythonWin programs.  PythonWin is
available as part of pywin32 project and as a part of the ActivePython
distribution.

Eric is an IDE built on PyQt and the Scintilla editing component.

trepan3k is a gdb-like debugger.

Visual Studio Code is an IDE with debugging tools that integrates with
version-control software.

There are a number of commercial Python IDEs that include graphical
debuggers. They include:

* Wing IDE

* Komodo IDE

* PyCharm


Are there tools to help find bugs or perform static analysis?
-------------------------------------------------------------

Yes.

Pylint and Pyflakes do basic checking that will help you catch bugs
sooner.

Static type checkers such as Mypy, Pyre, and Pytype can check type
hints in Python source code.


How can I create a stand-alone binary from a Python script?
-----------------------------------------------------------

You don’t need the ability to compile Python to C code if all you want
is a stand-alone program that users can download and run without
having to install the Python distribution first.  There are a number
of tools that determine the set of modules required by a program and
bind these modules together with a Python binary to produce a single
executable.

One is to use the freeze tool, which is included in the Python source
tree as Tools/freeze. It converts Python byte code to C arrays; with a
C compiler you can embed all your modules into a new program, which is
then linked with the standard Python modules.

It works by scanning your source recursively for import statements (in
both forms) and looking for the modules in the standard Python path as
well as in the source directory (for built-in modules).  It then turns
the bytecode for modules written in Python into C code (array
initializers that can be turned into code objects using the marshal
module) and creates a custom-made config file that only contains those
built-in modules which are actually used in the program.  It then
compiles the generated C code and links it with the rest of the Python
interpreter to form a self-contained binary which acts exactly like
your script.

The following packages can help with the creation of console and GUI
executables:

* Nuitka (Cross-platform)

* PyInstaller (Cross-platform)

* PyOxidizer (Cross-platform)

* cx_Freeze (Cross-platform)

* py2app (macOS only)

* py2exe (Windows only)


Are there coding standards or a style guide for Python programs?
----------------------------------------------------------------

Yes.  The coding style required for standard library modules is
documented as **PEP 8**.


Core Language
=============


Why am I getting an UnboundLocalError when the variable has a value?
--------------------------------------------------------------------

It can be a surprise to get the "UnboundLocalError" in previously
working code when it is modified by adding an assignment statement
somewhere in the body of a function.

This code:

>>> x = 10
>>> def bar():
...     print(x)
...
>>> bar()
10

works, but this code:

>>> x = 10
>>> def foo():
...     print(x)
...     x += 1

results in an "UnboundLocalError":

>>> foo()
Traceback (most recent call last):
  ...
UnboundLocalError: local variable 'x' referenced before assignment

This is because when you make an assignment to a variable in a scope,
that variable becomes local to that scope and shadows any similarly
named variable in the outer scope.  Since the last statement in foo
assigns a new value to "x", the compiler recognizes it as a local
variable.  Consequently when the earlier "print(x)" attempts to print
the uninitialized local variable and an error results.

In the example above you can access the outer scope variable by
declaring it global:

>>> x = 10
>>> def foobar():
...     global x
...     print(x)
...     x += 1
...
>>> foobar()
10

This explicit declaration is required in order to remind you that
(unlike the superficially analogous situation with class and instance
variables) you are actually modifying the value of the variable in the
outer scope:

>>> print(x)
11

You can do a similar thing in a nested scope using the "nonlocal"
keyword:

>>> def foo():
...    x = 10
...    def bar():
...        nonlocal x
...        print(x)
...        x += 1
...    bar()
...    print(x)
...
>>> foo()
10
11


What are the rules for local and global variables in Python?
------------------------------------------------------------

In Python, variables that are only referenced inside a function are
implicitly global.  If a variable is assigned a value anywhere within
the function’s body, it’s assumed to be a local unless explicitly
declared as global.

Though a bit surprising at first, a moment’s consideration explains
this.  On one hand, requiring "global" for assigned variables provides
a bar against unintended side-effects.  On the other hand, if "global"
was required for all global references, you’d be using "global" all
the time.  You’d have to declare as global every reference to a built-
in function or to a component of an imported module.  This clutter
would defeat the usefulness of the "global" declaration for
identifying side-effects.


Why do lambdas defined in a loop with different values all return the same result?
----------------------------------------------------------------------------------

Assume you use a for loop to define a few different lambdas (or even
plain functions), e.g.:

   >>> squares = []
   >>> for x in range(5):
   ...     squares.append(lambda: x**2)

This gives you a list that contains 5 lambdas that calculate "x**2".
You might expect that, when called, they would return, respectively,
"0", "1", "4", "9", and "16".  However, when you actually try you will
see that they all return "16":

   >>> squares[2]()
   16
   >>> squares[4]()
   16

This happens because "x" is not local to the lambdas, but is defined
in the outer scope, and it is accessed when the lambda is called — not
when it is defined.  At the end of the loop, the value of "x" is "4",
so all the functions now return "4**2", i.e. "16".  You can also
verify this by changing the value of "x" and see how the results of
the lambdas change:

   >>> x = 8
   >>> squares[2]()
   64

In order to avoid this, you need to save the values in variables local
to the lambdas, so that they don’t rely on the value of the global
"x":

   >>> squares = []
   >>> for x in range(5):
   ...     squares.append(lambda n=x: n**2)

Here, "n=x" creates a new variable "n" local to the lambda and
computed when the lambda is defined so that it has the same value that
"x" had at that point in the loop.  This means that the value of "n"
will be "0" in the first lambda, "1" in the second, "2" in the third,
and so on. Therefore each lambda will now return the correct result:

   >>> squares[2]()
   4
   >>> squares[4]()
   16

Note that this behaviour is not peculiar to lambdas, but applies to
regular functions too.


How do I share global variables across modules?
-----------------------------------------------

The canonical way to share information across modules within a single
program is to create a special module (often called config or cfg).
Just import the config module in all modules of your application; the
module then becomes available as a global name.  Because there is only
one instance of each module, any changes made to the module object get
reflected everywhere.  For example:

config.py:

   x = 0   # Default value of the 'x' configuration setting

mod.py:

   import config
   config.x = 1

main.py:

   import config
   import mod
   print(config.x)

Note that using a module is also the basis for implementing the
singleton design pattern, for the same reason.


What are the “best practices” for using import in a module?
-----------------------------------------------------------

In general, don’t use "from modulename import *".  Doing so clutters
the importer’s namespace, and makes it much harder for linters to
detect undefined names.

Import modules at the top of a file.  Doing so makes it clear what
other modules your code requires and avoids questions of whether the
module name is in scope. Using one import per line makes it easy to
add and delete module imports, but using multiple imports per line
uses less screen space.

It’s good practice if you import modules in the following order:

1. standard library modules – e.g. "sys", "os", "argparse", "re"

2. third-party library modules (anything installed in Python’s site-
   packages directory) – e.g. "dateutil", "requests", "PIL.Image"

3. locally developed modules

It is sometimes necessary to move imports to a function or class to
avoid problems with circular imports.  Gordon McMillan says:

   Circular imports are fine where both modules use the “import
   <module>” form of import.  They fail when the 2nd module wants to
   grab a name out of the first (“from module import name”) and the
   import is at the top level.  That’s because names in the 1st are
   not yet available, because the first module is busy importing the
   2nd.

In this case, if the second module is only used in one function, then
the import can easily be moved into that function.  By the time the
import is called, the first module will have finished initializing,
and the second module can do its import.

It may also be necessary to move imports out of the top level of code
if some of the modules are platform-specific.  In that case, it may
not even be possible to import all of the modules at the top of the
file.  In this case, importing the correct modules in the
corresponding platform-specific code is a good option.

Only move imports into a local scope, such as inside a function
definition, if it’s necessary to solve a problem such as avoiding a
circular import or are trying to reduce the initialization time of a
module.  This technique is especially helpful if many of the imports
are unnecessary depending on how the program executes.  You may also
want to move imports into a function if the modules are only ever used
in that function.  Note that loading a module the first time may be
expensive because of the one time initialization of the module, but
loading a module multiple times is virtually free, costing only a
couple of dictionary lookups.  Even if the module name has gone out of
scope, the module is probably available in "sys.modules".


Why are default values shared between objects?
----------------------------------------------

This type of bug commonly bites neophyte programmers.  Consider this
function:

   def foo(mydict={}):  # Danger: shared reference to one dict for all calls
       ... compute something ...
       mydict[key] = value
       return mydict

The first time you call this function, "mydict" contains a single
item.  The second time, "mydict" contains two items because when
"foo()" begins executing, "mydict" starts out with an item already in
it.

It is often expected that a function call creates new objects for
default values. This is not what happens. Default values are created
exactly once, when the function is defined.  If that object is
changed, like the dictionary in this example, subsequent calls to the
function will refer to this changed object.

By definition, immutable objects such as numbers, strings, tuples, and
"None", are safe from change. Changes to mutable objects such as
dictionaries, lists, and class instances can lead to confusion.

Because of this feature, it is good programming practice to not use
mutable objects as default values.  Instead, use "None" as the default
value and inside the function, check if the parameter is "None" and
create a new list/dictionary/whatever if it is.  For example, don’t
write:

   def foo(mydict={}):
       ...

but:

   def foo(mydict=None):
       if mydict is None:
           mydict = {}  # create a new dict for local namespace

This feature can be useful.  When you have a function that’s time-
consuming to compute, a common technique is to cache the parameters
and the resulting value of each call to the function, and return the
cached value if the same value is requested again.  This is called
“memoizing”, and can be implemented like this:

   # Callers can only provide two parameters and optionally pass _cache by keyword
   def expensive(arg1, arg2, *, _cache={}):
       if (arg1, arg2) in _cache:
           return _cache[(arg1, arg2)]

       # Calculate the value
       result = ... expensive computation ...
       _cache[(arg1, arg2)] = result           # Store result in the cache
       return result

You could use a global variable containing a dictionary instead of the
default value; it’s a matter of taste.


How can I pass optional or keyword parameters from one function to another?
---------------------------------------------------------------------------

Collect the arguments using the "*" and "**" specifiers in the
function’s parameter list; this gives you the positional arguments as
a tuple and the keyword arguments as a dictionary.  You can then pass
these arguments when calling another function by using "*" and "**":

   def f(x, *args, **kwargs):
       ...
       kwargs['width'] = '14.3c'
       ...
       g(x, *args, **kwargs)


What is the difference between arguments and parameters?
--------------------------------------------------------

*Parameters* are defined by the names that appear in a function
definition, whereas *arguments* are the values actually passed to a
function when calling it.  Parameters define what *kind of arguments*
a function can accept.  For example, given the function definition:

   def func(foo, bar=None, **kwargs):
       pass

*foo*, *bar* and *kwargs* are parameters of "func".  However, when
calling "func", for example:

   func(42, bar=314, extra=somevar)

the values "42", "314", and "somevar" are arguments.


Why did changing list ‘y’ also change list ‘x’?
-----------------------------------------------

If you wrote code like:

   >>> x = []
   >>> y = x
   >>> y.append(10)
   >>> y
   [10]
   >>> x
   [10]

you might be wondering why appending an element to "y" changed "x"
too.

There are two factors that produce this result:

1. Variables are simply names that refer to objects.  Doing "y = x"
   doesn’t create a copy of the list – it creates a new variable "y"
   that refers to the same object "x" refers to.  This means that
   there is only one object (the list), and both "x" and "y" refer to
   it.

2. Lists are *mutable*, which means that you can change their content.

After the call to "append()", the content of the mutable object has
changed from "[]" to "[10]".  Since both the variables refer to the
same object, using either name accesses the modified value "[10]".

If we instead assign an immutable object to "x":

   >>> x = 5  # ints are immutable
   >>> y = x
   >>> x = x + 1  # 5 can't be mutated, we are creating a new object here
   >>> x
   6
   >>> y
   5

we can see that in this case "x" and "y" are not equal anymore.  This
is because integers are *immutable*, and when we do "x = x + 1" we are
not mutating the int "5" by incrementing its value; instead, we are
creating a new object (the int "6") and assigning it to "x" (that is,
changing which object "x" refers to).  After this assignment we have
two objects (the ints "6" and "5") and two variables that refer to
them ("x" now refers to "6" but "y" still refers to "5").

Some operations (for example "y.append(10)" and "y.sort()") mutate the
object, whereas superficially similar operations (for example "y = y +
[10]" and "sorted(y)") create a new object.  In general in Python (and
in all cases in the standard library) a method that mutates an object
will return "None" to help avoid getting the two types of operations
confused.  So if you mistakenly write "y.sort()" thinking it will give
you a sorted copy of "y", you’ll instead end up with "None", which
will likely cause your program to generate an easily diagnosed error.

However, there is one class of operations where the same operation
sometimes has different behaviors with different types:  the augmented
assignment operators.  For example, "+=" mutates lists but not tuples
or ints ("a_list += [1, 2, 3]" is equivalent to "a_list.extend([1, 2,
3])" and mutates "a_list", whereas "some_tuple += (1, 2, 3)" and
"some_int += 1" create new objects).

In other words:

* If we have a mutable object ("list", "dict", "set", etc.), we can
  use some specific operations to mutate it and all the variables that
  refer to it will see the change.

* If we have an immutable object ("str", "int", "tuple", etc.), all
  the variables that refer to it will always see the same value, but
  operations that transform that value into a new value always return
  a new object.

If you want to know if two variables refer to the same object or not,
you can use the "is" operator, or the built-in function "id()".


How do I write a function with output parameters (call by reference)?
---------------------------------------------------------------------

Remember that arguments are passed by assignment in Python.  Since
assignment just creates references to objects, there’s no alias
between an argument name in the caller and callee, and so no call-by-
reference per se.  You can achieve the desired effect in a number of
ways.

1. By returning a tuple of the results:

      >>> def func1(a, b):
      ...     a = 'new-value'        # a and b are local names
      ...     b = b + 1              # assigned to new objects
      ...     return a, b            # return new values
      ...
      >>> x, y = 'old-value', 99
      >>> func1(x, y)
      ('new-value', 100)

   This is almost always the clearest solution.

2. By using global variables.  This isn’t thread-safe, and is not
   recommended.

3. By passing a mutable (changeable in-place) object:

      >>> def func2(a):
      ...     a[0] = 'new-value'     # 'a' references a mutable list
      ...     a[1] = a[1] + 1        # changes a shared object
      ...
      >>> args = ['old-value', 99]
      >>> func2(args)
      >>> args
      ['new-value', 100]

4. By passing in a dictionary that gets mutated:

      >>> def func3(args):
      ...     args['a'] = 'new-value'     # args is a mutable dictionary
      ...     args['b'] = args['b'] + 1   # change it in-place
      ...
      >>> args = {'a': 'old-value', 'b': 99}
      >>> func3(args)
      >>> args
      {'a': 'new-value', 'b': 100}

5. Or bundle up values in a class instance:

      >>> class Namespace:
      ...     def __init__(self, /, **args):
      ...         for key, value in args.items():
      ...             setattr(self, key, value)
      ...
      >>> def func4(args):
      ...     args.a = 'new-value'        # args is a mutable Namespace
      ...     args.b = args.b + 1         # change object in-place
      ...
      >>> args = Namespace(a='old-value', b=99)
      >>> func4(args)
      >>> vars(args)
      {'a': 'new-value', 'b': 100}

   There’s almost never a good reason to get this complicated.

Your best choice is to return a tuple containing the multiple results.


How do you make a higher order function in Python?
--------------------------------------------------

You have two choices: you can use nested scopes or you can use
callable objects. For example, suppose you wanted to define
"linear(a,b)" which returns a function "f(x)" that computes the value
"a*x+b".  Using nested scopes:

   def linear(a, b):
       def result(x):
           return a * x + b
       return result

Or using a callable object:

   class linear:

       def __init__(self, a, b):
           self.a, self.b = a, b

       def __call__(self, x):
           return self.a * x + self.b

In both cases,

   taxes = linear(0.3, 2)

gives a callable object where "taxes(10e6) == 0.3 * 10e6 + 2".

The callable object approach has the disadvantage that it is a bit
slower and results in slightly longer code.  However, note that a
collection of callables can share their signature via inheritance:

   class exponential(linear):
       # __init__ inherited
       def __call__(self, x):
           return self.a * (x ** self.b)

Object can encapsulate state for several methods:

   class counter:

       value = 0

       def set(self, x):
           self.value = x

       def up(self):
           self.value = self.value + 1

       def down(self):
           self.value = self.value - 1

   count = counter()
   inc, dec, reset = count.up, count.down, count.set

Here "inc()", "dec()" and "reset()" act like functions which share the
same counting variable.


How do I copy an object in Python?
----------------------------------

In general, try "copy.copy()" or "copy.deepcopy()" for the general
case. Not all objects can be copied, but most can.

Some objects can be copied more easily.  Dictionaries have a "copy()"
method:

   newdict = olddict.copy()

Sequences can be copied by slicing:

   new_l = l[:]


How can I find the methods or attributes of an object?
------------------------------------------------------

For an instance "x" of a user-defined class, "dir(x)" returns an
alphabetized list of the names containing the instance attributes and
methods and attributes defined by its class.


How can my code discover the name of an object?
-----------------------------------------------

Generally speaking, it can’t, because objects don’t really have names.
Essentially, assignment always binds a name to a value; the same is
true of "def" and "class" statements, but in that case the value is a
callable. Consider the following code:

   >>> class A:
   ...     pass
   ...
   >>> B = A
   >>> a = B()
   >>> b = a
   >>> print(b)
   <__main__.A object at 0x16D07CC>
   >>> print(a)
   <__main__.A object at 0x16D07CC>

Arguably the class has a name: even though it is bound to two names
and invoked through the name "B" the created instance is still
reported as an instance of class "A".  However, it is impossible to
say whether the instance’s name is "a" or "b", since both names are
bound to the same value.

Generally speaking it should not be necessary for your code to “know
the names” of particular values. Unless you are deliberately writing
introspective programs, this is usually an indication that a change of
approach might be beneficial.

In comp.lang.python, Fredrik Lundh once gave an excellent analogy in
answer to this question:

   The same way as you get the name of that cat you found on your
   porch: the cat (object) itself cannot tell you its name, and it
   doesn’t really care – so the only way to find out what it’s called
   is to ask all your neighbours (namespaces) if it’s their cat
   (object)…

   ….and don’t be surprised if you’ll find that it’s known by many
   names, or no name at all!


What’s up with the comma operator’s precedence?
-----------------------------------------------

Comma is not an operator in Python.  Consider this session:

   >>> "a" in "b", "a"
   (False, 'a')

Since the comma is not an operator, but a separator between
expressions the above is evaluated as if you had entered:

   ("a" in "b"), "a"

not:

   "a" in ("b", "a")

The same is true of the various assignment operators ("=", "+=" etc).
They are not truly operators but syntactic delimiters in assignment
statements.


Is there an equivalent of C’s “?:” ternary operator?
----------------------------------------------------

Yes, there is. The syntax is as follows:

   [on_true] if [expression] else [on_false]

   x, y = 50, 25
   small = x if x < y else y

Before this syntax was introduced in Python 2.5, a common idiom was to
use logical operators:

   [expression] and [on_true] or [on_false]

However, this idiom is unsafe, as it can give wrong results when
*on_true* has a false boolean value.  Therefore, it is always better
to use the "... if ... else ..." form.


Is it possible to write obfuscated one-liners in Python?
--------------------------------------------------------

Yes.  Usually this is done by nesting "lambda" within "lambda".  See
the following three examples, slightly adapted from Ulf Bartelt:

   from functools import reduce

   # Primes < 1000
   print(list(filter(None,map(lambda y:y*reduce(lambda x,y:x*y!=0,
   map(lambda x,y=y:y%x,range(2,int(pow(y,0.5)+1))),1),range(2,1000)))))

   # First 10 Fibonacci numbers
   print(list(map(lambda x,f=lambda x,f:(f(x-1,f)+f(x-2,f)) if x>1 else 1:
   f(x,f), range(10))))

   # Mandelbrot set
   print((lambda Ru,Ro,Iu,Io,IM,Sx,Sy:reduce(lambda x,y:x+'\n'+y,map(lambda y,
   Iu=Iu,Io=Io,Ru=Ru,Ro=Ro,Sy=Sy,L=lambda yc,Iu=Iu,Io=Io,Ru=Ru,Ro=Ro,i=IM,
   Sx=Sx,Sy=Sy:reduce(lambda x,y:x+y,map(lambda x,xc=Ru,yc=yc,Ru=Ru,Ro=Ro,
   i=i,Sx=Sx,F=lambda xc,yc,x,y,k,f=lambda xc,yc,x,y,k,f:(k<=0)or (x*x+y*y
   >=4.0) or 1+f(xc,yc,x*x-y*y+xc,2.0*x*y+yc,k-1,f):f(xc,yc,x,y,k,f):chr(
   64+F(Ru+x*(Ro-Ru)/Sx,yc,0,0,i)),range(Sx))):L(Iu+y*(Io-Iu)/Sy),range(Sy
   ))))(-2.1, 0.7, -1.2, 1.2, 30, 80, 24))
   #    \___ ___/  \___ ___/  |   |   |__ lines on screen
   #        V          V      |   |______ columns on screen
   #        |          |      |__________ maximum of "iterations"
   #        |          |_________________ range on y axis
   #        |____________________________ range on x axis

Don’t try this at home, kids!


What does the slash(/) in the parameter list of a function mean?
----------------------------------------------------------------

A slash in the argument list of a function denotes that the parameters
prior to it are positional-only.  Positional-only parameters are the
ones without an externally usable name.  Upon calling a function that
accepts positional-only parameters, arguments are mapped to parameters
based solely on their position. For example, "divmod()" is a function
that accepts positional-only parameters. Its documentation looks like
this:

   >>> help(divmod)
   Help on built-in function divmod in module builtins:

   divmod(x, y, /)
       Return the tuple (x//y, x%y).  Invariant: div*y + mod == x.

The slash at the end of the parameter list means that both parameters
are positional-only. Thus, calling "divmod()" with keyword arguments
would lead to an error:

   >>> divmod(x=3, y=4)
   Traceback (most recent call last):
     File "<stdin>", line 1, in <module>
   TypeError: divmod() takes no keyword arguments


Numbers and strings
===================


How do I specify hexadecimal and octal integers?
------------------------------------------------

To specify an octal digit, precede the octal value with a zero, and
then a lower or uppercase “o”.  For example, to set the variable “a”
to the octal value “10” (8 in decimal), type:

   >>> a = 0o10
   >>> a
   8

Hexadecimal is just as easy.  Simply precede the hexadecimal number
with a zero, and then a lower or uppercase “x”.  Hexadecimal digits
can be specified in lower or uppercase.  For example, in the Python
interpreter:

   >>> a = 0xa5
   >>> a
   165
   >>> b = 0XB2
   >>> b
   178


Why does -22 // 10 return -3?
-----------------------------

It’s primarily driven by the desire that "i % j" have the same sign as
"j". If you want that, and also want:

   i == (i // j) * j + (i % j)

then integer division has to return the floor.  C also requires that
identity to hold, and then compilers that truncate "i // j" need to
make "i % j" have the same sign as "i".

There are few real use cases for "i % j" when "j" is negative.  When
"j" is positive, there are many, and in virtually all of them it’s
more useful for "i % j" to be ">= 0".  If the clock says 10 now, what
did it say 200 hours ago?  "-190 % 12 == 2" is useful; "-190 % 12 ==
-10" is a bug waiting to bite.


How do I get int literal attribute instead of SyntaxError?
----------------------------------------------------------

Trying to lookup an "int" literal attribute in the normal manner gives
a "SyntaxError" because the period is seen as a decimal point:

   >>> 1.__class__
     File "<stdin>", line 1
     1.__class__
      ^
   SyntaxError: invalid decimal literal

The solution is to separate the literal from the period with either a
space or parentheses.

>>> 1 .__class__
<class 'int'>
>>> (1).__class__
<class 'int'>


How do I convert a string to a number?
--------------------------------------

For integers, use the built-in "int()" type constructor, e.g.
"int('144') == 144".  Similarly, "float()" converts to a floating-
point number, e.g. "float('144') == 144.0".

By default, these interpret the number as decimal, so that
"int('0144') == 144" holds true, and "int('0x144')" raises
"ValueError". "int(string, base)" takes the base to convert from as a
second optional argument, so "int( '0x144', 16) == 324".  If the base
is specified as 0, the number is interpreted using Python’s rules: a
leading ‘0o’ indicates octal, and ‘0x’ indicates a hex number.

Do not use the built-in function "eval()" if all you need is to
convert strings to numbers.  "eval()" will be significantly slower and
it presents a security risk: someone could pass you a Python
expression that might have unwanted side effects.  For example,
someone could pass "__import__('os').system("rm -rf $HOME")" which
would erase your home directory.

"eval()" also has the effect of interpreting numbers as Python
expressions, so that e.g. "eval('09')" gives a syntax error because
Python does not allow leading ‘0’ in a decimal number (except ‘0’).


How do I convert a number to a string?
--------------------------------------

To convert, e.g., the number "144" to the string "'144'", use the
built-in type constructor "str()".  If you want a hexadecimal or octal
representation, use the built-in functions "hex()" or "oct()".  For
fancy formatting, see the f-strings and Format String Syntax sections,
e.g. ""{:04d}".format(144)" yields "'0144'" and
""{:.3f}".format(1.0/3.0)" yields "'0.333'".


How do I modify a string in place?
----------------------------------

You can’t, because strings are immutable.  In most situations, you
should simply construct a new string from the various parts you want
to assemble it from.  However, if you need an object with the ability
to modify in-place unicode data, try using an "io.StringIO" object or
the "array" module:

   >>> import io
   >>> s = "Hello, world"
   >>> sio = io.StringIO(s)
   >>> sio.getvalue()
   'Hello, world'
   >>> sio.seek(7)
   7
   >>> sio.write("there!")
   6
   >>> sio.getvalue()
   'Hello, there!'

   >>> import array
   >>> a = array.array('w', s)
   >>> print(a)
   array('w', 'Hello, world')
   >>> a[0] = 'y'
   >>> print(a)
   array('w', 'yello, world')
   >>> a.tounicode()
   'yello, world'


How do I use strings to call functions/methods?
-----------------------------------------------

There are various techniques.

* The best is to use a dictionary that maps strings to functions.  The
  primary advantage of this technique is that the strings do not need
  to match the names of the functions.  This is also the primary
  technique used to emulate a case construct:

     def a():
         pass

     def b():
         pass

     dispatch = {'go': a, 'stop': b}  # Note lack of parens for funcs

     dispatch[get_input()]()  # Note trailing parens to call function

* Use the built-in function "getattr()":

     import foo
     getattr(foo, 'bar')()

  Note that "getattr()" works on any object, including classes, class
  instances, modules, and so on.

  This is used in several places in the standard library, like this:

     class Foo:
         def do_foo(self):
             ...

         def do_bar(self):
             ...

     f = getattr(foo_instance, 'do_' + opname)
     f()

* Use "locals()" to resolve the function name:

     def myFunc():
         print("hello")

     fname = "myFunc"

     f = locals()[fname]
     f()


Is there an equivalent to Perl’s "chomp()" for removing trailing newlines from strings?
---------------------------------------------------------------------------------------

You can use "S.rstrip("\r\n")" to remove all occurrences of any line
terminator from the end of the string "S" without removing other
trailing whitespace.  If the string "S" represents more than one line,
with several empty lines at the end, the line terminators for all the
blank lines will be removed:

   >>> lines = ("line 1 \r\n"
   ...          "\r\n"
   ...          "\r\n")
   >>> lines.rstrip("\n\r")
   'line 1 '

Since this is typically only desired when reading text one line at a
time, using "S.rstrip()" this way works well.


Is there a "scanf()" or "sscanf()" equivalent?
----------------------------------------------

Not as such.

For simple input parsing, the easiest approach is usually to split the
line into whitespace-delimited words using the "split()" method of
string objects and then convert decimal strings to numeric values
using "int()" or "float()".  "split()" supports an optional “sep”
parameter which is useful if the line uses something other than
whitespace as a separator.

For more complicated input parsing, regular expressions are more
powerful than C’s "sscanf" and better suited for the task.


What does "UnicodeDecodeError" or "UnicodeEncodeError" error mean?
------------------------------------------------------------------

See the Unicode HOWTO.


Can I end a raw string with an odd number of backslashes?
---------------------------------------------------------

A raw string ending with an odd number of backslashes will escape the
string’s quote:

   >>> r'C:\this\will\not\work\'
     File "<stdin>", line 1
       r'C:\this\will\not\work\'
       ^
   SyntaxError: unterminated string literal (detected at line 1)

There are several workarounds for this. One is to use regular strings
and double the backslashes:

   >>> 'C:\\this\\will\\work\\'
   'C:\\this\\will\\work\\'

Another is to concatenate a regular string containing an escaped
backslash to the raw string:

   >>> r'C:\this\will\work' '\\'
   'C:\\this\\will\\work\\'

It is also possible to use "os.path.join()" to append a backslash on
Windows:

   >>> os.path.join(r'C:\this\will\work', '')
   'C:\\this\\will\\work\\'

Note that while a backslash will “escape” a quote for the purposes of
determining where the raw string ends, no escaping occurs when
interpreting the value of the raw string. That is, the backslash
remains present in the value of the raw string:

   >>> r'backslash\'preserved'
   "backslash\\'preserved"

Also see the specification in the language reference.


Performance
===========


My program is too slow. How do I speed it up?
---------------------------------------------

That’s a tough one, in general.  First, here are a list of things to
remember before diving further:

* Performance characteristics vary across Python implementations.
  This FAQ focuses on *CPython*.

* Behaviour can vary across operating systems, especially when talking
  about I/O or multi-threading.

* You should always find the hot spots in your program *before*
  attempting to optimize any code (see the "profile" module).

* Writing benchmark scripts will allow you to iterate quickly when
  searching for improvements (see the "timeit" module).

* It is highly recommended to have good code coverage (through unit
  testing or any other technique) before potentially introducing
  regressions hidden in sophisticated optimizations.

That being said, there are many tricks to speed up Python code.  Here
are some general principles which go a long way towards reaching
acceptable performance levels:

* Making your algorithms faster (or changing to faster ones) can yield
  much larger benefits than trying to sprinkle micro-optimization
  tricks all over your code.

* Use the right data structures.  Study documentation for the Built-in
  Types and the "collections" module.

* When the standard library provides a primitive for doing something,
  it is likely (although not guaranteed) to be faster than any
  alternative you may come up with.  This is doubly true for
  primitives written in C, such as builtins and some extension types.
  For example, be sure to use either the "list.sort()" built-in method
  or the related "sorted()" function to do sorting (and see the
  Sorting Techniques for examples of moderately advanced usage).

* Abstractions tend to create indirections and force the interpreter
  to work more.  If the levels of indirection outweigh the amount of
  useful work done, your program will be slower.  You should avoid
  excessive abstraction, especially under the form of tiny functions
  or methods (which are also often detrimental to readability).

If you have reached the limit of what pure Python can allow, there are
tools to take you further away.  For example, Cython can compile a
slightly modified version of Python code into a C extension, and can
be used on many different platforms.  Cython can take advantage of
compilation (and optional type annotations) to make your code
significantly faster than when interpreted.  If you are confident in
your C programming skills, you can also write a C extension module
yourself.

See also: The wiki page devoted to performance tips.


What is the most efficient way to concatenate many strings together?
--------------------------------------------------------------------

"str" and "bytes" objects are immutable, therefore concatenating many
strings together is inefficient as each concatenation creates a new
object.  In the general case, the total runtime cost is quadratic in
the total string length.

To accumulate many "str" objects, the recommended idiom is to place
them into a list and call "str.join()" at the end:

   chunks = []
   for s in my_strings:
       chunks.append(s)
   result = ''.join(chunks)

(another reasonably efficient idiom is to use "io.StringIO")

To accumulate many "bytes" objects, the recommended idiom is to extend
a "bytearray" object using in-place concatenation (the "+=" operator):

   result = bytearray()
   for b in my_bytes_objects:
       result += b


Sequences (Tuples/Lists)
========================


How do I convert between tuples and lists?
------------------------------------------

The type constructor "tuple(seq)" converts any sequence (actually, any
iterable) into a tuple with the same items in the same order.

For example, "tuple([1, 2, 3])" yields "(1, 2, 3)" and "tuple('abc')"
yields "('a', 'b', 'c')".  If the argument is a tuple, it does not
make a copy but returns the same object, so it is cheap to call
"tuple()" when you aren’t sure that an object is already a tuple.

The type constructor "list(seq)" converts any sequence or iterable
into a list with the same items in the same order.  For example,
"list((1, 2, 3))" yields "[1, 2, 3]" and "list('abc')" yields "['a',
'b', 'c']".  If the argument is a list, it makes a copy just like
"seq[:]" would.


What’s a negative index?
------------------------

Python sequences are indexed with positive numbers and negative
numbers.  For positive numbers 0 is the first index 1 is the second
index and so forth.  For negative indices -1 is the last index and -2
is the penultimate (next to last) index and so forth.  Think of
"seq[-n]" as the same as "seq[len(seq)-n]".

Using negative indices can be very convenient.  For example "S[:-1]"
is all of the string except for its last character, which is useful
for removing the trailing newline from a string.


How do I iterate over a sequence in reverse order?
--------------------------------------------------

Use the "reversed()" built-in function:

   for x in reversed(sequence):
       ...  # do something with x ...

This won’t touch your original sequence, but build a new copy with
reversed order to iterate over.


How do you remove duplicates from a list?
-----------------------------------------

See the Python Cookbook for a long discussion of many ways to do this:

   https://code.activestate.com/recipes/52560/

If you don’t mind reordering the list, sort it and then scan from the
end of the list, deleting duplicates as you go:

   if mylist:
       mylist.sort()
       last = mylist[-1]
       for i in range(len(mylist)-2, -1, -1):
           if last == mylist[i]:
               del mylist[i]
           else:
               last = mylist[i]

If all elements of the list may be used as set keys (i.e. they are all
*hashable*) this is often faster

   mylist = list(set(mylist))

This converts the list into a set, thereby removing duplicates, and
then back into a list.


How do you remove multiple items from a list
--------------------------------------------

As with removing duplicates, explicitly iterating in reverse with a
delete condition is one possibility.  However, it is easier and faster
to use slice replacement with an implicit or explicit forward
iteration. Here are three variations.:

   mylist[:] = filter(keep_function, mylist)
   mylist[:] = (x for x in mylist if keep_condition)
   mylist[:] = [x for x in mylist if keep_condition]

The list comprehension may be fastest.


How do you make an array in Python?
-----------------------------------

Use a list:

   ["this", 1, "is", "an", "array"]

Lists are equivalent to C or Pascal arrays in their time complexity;
the primary difference is that a Python list can contain objects of
many different types.

The "array" module also provides methods for creating arrays of fixed
types with compact representations, but they are slower to index than
lists.  Also note that NumPy and other third party packages define
array-like structures with various characteristics as well.

To get Lisp-style linked lists, you can emulate *cons cells* using
tuples:

   lisp_list = ("like",  ("this",  ("example", None) ) )

If mutability is desired, you could use lists instead of tuples.  Here
the analogue of a Lisp *car* is "lisp_list[0]" and the analogue of
*cdr* is "lisp_list[1]".  Only do this if you’re sure you really need
to, because it’s usually a lot slower than using Python lists.


How do I create a multidimensional list?
----------------------------------------

You probably tried to make a multidimensional array like this:

   >>> A = [[None] * 2] * 3

This looks correct if you print it:

   >>> A
   [[None, None], [None, None], [None, None]]

But when you assign a value, it shows up in multiple places:

   >>> A[0][0] = 5
   >>> A
   [[5, None], [5, None], [5, None]]

The reason is that replicating a list with "*" doesn’t create copies,
it only creates references to the existing objects.  The "*3" creates
a list containing 3 references to the same list of length two.
Changes to one row will show in all rows, which is almost certainly
not what you want.

The suggested approach is to create a list of the desired length first
and then fill in each element with a newly created list:

   A = [None] * 3
   for i in range(3):
       A[i] = [None] * 2

This generates a list containing 3 different lists of length two.  You
can also use a list comprehension:

   w, h = 2, 3
   A = [[None] * w for i in range(h)]

Or, you can use an extension that provides a matrix datatype; NumPy is
the best known.


How do I apply a method or function to a sequence of objects?
-------------------------------------------------------------

To call a method or function and accumulate the return values is a
list, a *list comprehension* is an elegant solution:

   result = [obj.method() for obj in mylist]

   result = [function(obj) for obj in mylist]

To just run the method or function without saving the return values, a
plain "for" loop will suffice:

   for obj in mylist:
       obj.method()

   for obj in mylist:
       function(obj)


Why does a_tuple[i] += [‘item’] raise an exception when the addition works?
---------------------------------------------------------------------------

This is because of a combination of the fact that augmented assignment
operators are *assignment* operators, and the difference between
mutable and immutable objects in Python.

This discussion applies in general when augmented assignment operators
are applied to elements of a tuple that point to mutable objects, but
we’ll use a "list" and "+=" as our exemplar.

If you wrote:

   >>> a_tuple = (1, 2)
   >>> a_tuple[0] += 1
   Traceback (most recent call last):
      ...
   TypeError: 'tuple' object does not support item assignment

The reason for the exception should be immediately clear: "1" is added
to the object "a_tuple[0]" points to ("1"), producing the result
object, "2", but when we attempt to assign the result of the
computation, "2", to element "0" of the tuple, we get an error because
we can’t change what an element of a tuple points to.

Under the covers, what this augmented assignment statement is doing is
approximately this:

   >>> result = a_tuple[0] + 1
   >>> a_tuple[0] = result
   Traceback (most recent call last):
     ...
   TypeError: 'tuple' object does not support item assignment

It is the assignment part of the operation that produces the error,
since a tuple is immutable.

When you write something like:

   >>> a_tuple = (['foo'], 'bar')
   >>> a_tuple[0] += ['item']
   Traceback (most recent call last):
     ...
   TypeError: 'tuple' object does not support item assignment

The exception is a bit more surprising, and even more surprising is
the fact that even though there was an error, the append worked:

   >>> a_tuple[0]
   ['foo', 'item']

To see why this happens, you need to know that (a) if an object
implements an "__iadd__()" magic method, it gets called when the "+="
augmented assignment is executed, and its return value is what gets
used in the assignment statement; and (b) for lists, "__iadd__()" is
equivalent to calling "extend()" on the list and returning the list.
That’s why we say that for lists, "+=" is a “shorthand” for
"list.extend()":

   >>> a_list = []
   >>> a_list += [1]
   >>> a_list
   [1]

This is equivalent to:

   >>> result = a_list.__iadd__([1])
   >>> a_list = result

The object pointed to by a_list has been mutated, and the pointer to
the mutated object is assigned back to "a_list".  The end result of
the assignment is a no-op, since it is a pointer to the same object
that "a_list" was previously pointing to, but the assignment still
happens.

Thus, in our tuple example what is happening is equivalent to:

   >>> result = a_tuple[0].__iadd__(['item'])
   >>> a_tuple[0] = result
   Traceback (most recent call last):
     ...
   TypeError: 'tuple' object does not support item assignment

The "__iadd__()" succeeds, and thus the list is extended, but even
though "result" points to the same object that "a_tuple[0]" already
points to, that final assignment still results in an error, because
tuples are immutable.


I want to do a complicated sort: can you do a Schwartzian Transform in Python?
------------------------------------------------------------------------------

The technique, attributed to Randal Schwartz of the Perl community,
sorts the elements of a list by a metric which maps each element to
its “sort value”. In Python, use the "key" argument for the
"list.sort()" method:

   Isorted = L[:]
   Isorted.sort(key=lambda s: int(s[10:15]))


How can I sort one list by values from another list?
----------------------------------------------------

Merge them into an iterator of tuples, sort the resulting list, and
then pick out the element you want.

   >>> list1 = ["what", "I'm", "sorting", "by"]
   >>> list2 = ["something", "else", "to", "sort"]
   >>> pairs = zip(list1, list2)
   >>> pairs = sorted(pairs)
   >>> pairs
   [("I'm", 'else'), ('by', 'sort'), ('sorting', 'to'), ('what', 'something')]
   >>> result = [x[1] for x in pairs]
   >>> result
   ['else', 'sort', 'to', 'something']


Objects
=======


What is a class?
----------------

A class is the particular object type created by executing a class
statement. Class objects are used as templates to create instance
objects, which embody both the data (attributes) and code (methods)
specific to a datatype.

A class can be based on one or more other classes, called its base
class(es). It then inherits the attributes and methods of its base
classes. This allows an object model to be successively refined by
inheritance.  You might have a generic "Mailbox" class that provides
basic accessor methods for a mailbox, and subclasses such as
"MboxMailbox", "MaildirMailbox", "OutlookMailbox" that handle various
specific mailbox formats.


What is a method?
-----------------

A method is a function on some object "x" that you normally call as
"x.name(arguments...)".  Methods are defined as functions inside the
class definition:

   class C:
       def meth(self, arg):
           return arg * 2 + self.attribute


What is self?
-------------

Self is merely a conventional name for the first argument of a method.
A method defined as "meth(self, a, b, c)" should be called as
"x.meth(a, b, c)" for some instance "x" of the class in which the
definition occurs; the called method will think it is called as
"meth(x, a, b, c)".

See also Why must ‘self’ be used explicitly in method definitions and
calls?.


How do I check if an object is an instance of a given class or of a subclass of it?
-----------------------------------------------------------------------------------

Use the built-in function "isinstance(obj, cls)".  You can check if an
object is an instance of any of a number of classes by providing a
tuple instead of a single class, e.g. "isinstance(obj, (class1,
class2, ...))", and can also check whether an object is one of
Python’s built-in types, e.g. "isinstance(obj, str)" or
"isinstance(obj, (int, float, complex))".

Note that "isinstance()" also checks for virtual inheritance from an
*abstract base class*.  So, the test will return "True" for a
registered class even if hasn’t directly or indirectly inherited from
it.  To test for “true inheritance”, scan the *MRO* of the class:

   from collections.abc import Mapping

   class P:
        pass

   class C(P):
       pass

   Mapping.register(P)

   >>> c = C()
   >>> isinstance(c, C)        # direct
   True
   >>> isinstance(c, P)        # indirect
   True
   >>> isinstance(c, Mapping)  # virtual
   True

   # Actual inheritance chain
   >>> type(c).__mro__
   (<class 'C'>, <class 'P'>, <class 'object'>)

   # Test for "true inheritance"
   >>> Mapping in type(c).__mro__
   False

Note that most programs do not use "isinstance()" on user-defined
classes very often.  If you are developing the classes yourself, a
more proper object-oriented style is to define methods on the classes
that encapsulate a particular behaviour, instead of checking the
object’s class and doing a different thing based on what class it is.
For example, if you have a function that does something:

   def search(obj):
       if isinstance(obj, Mailbox):
           ...  # code to search a mailbox
       elif isinstance(obj, Document):
           ...  # code to search a document
       elif ...

A better approach is to define a "search()" method on all the classes
and just call it:

   class Mailbox:
       def search(self):
           ...  # code to search a mailbox

   class Document:
       def search(self):
           ...  # code to search a document

   obj.search()


What is delegation?
-------------------

Delegation is an object oriented technique (also called a design
pattern). Let’s say you have an object "x" and want to change the
behaviour of just one of its methods.  You can create a new class that
provides a new implementation of the method you’re interested in
changing and delegates all other methods to the corresponding method
of "x".

Python programmers can easily implement delegation.  For example, the
following class implements a class that behaves like a file but
converts all written data to uppercase:

   class UpperOut:

       def __init__(self, outfile):
           self._outfile = outfile

       def write(self, s):
           self._outfile.write(s.upper())

       def __getattr__(self, name):
           return getattr(self._outfile, name)

Here the "UpperOut" class redefines the "write()" method to convert
the argument string to uppercase before calling the underlying
"self._outfile.write()" method.  All other methods are delegated to
the underlying "self._outfile" object.  The delegation is accomplished
via the "__getattr__()" method; consult the language reference for
more information about controlling attribute access.

Note that for more general cases delegation can get trickier. When
attributes must be set as well as retrieved, the class must define a
"__setattr__()" method too, and it must do so carefully.  The basic
implementation of "__setattr__()" is roughly equivalent to the
following:

   class X:
       ...
       def __setattr__(self, name, value):
           self.__dict__[name] = value
       ...

Many "__setattr__()" implementations call "object.__setattr__()" to
set an attribute on self without causing infinite recursion:

   class X:
       def __setattr__(self, name, value):
           # Custom logic here...
           object.__setattr__(self, name, value)

Alternatively, it is possible to set attributes by inserting entries
into "self.__dict__" directly.


How do I call a method defined in a base class from a derived class that extends it?
------------------------------------------------------------------------------------

Use the built-in "super()" function:

   class Derived(Base):
       def meth(self):
           super().meth()  # calls Base.meth

In the example, "super()" will automatically determine the instance
from which it was called (the "self" value), look up the *method
resolution order* (MRO) with "type(self).__mro__", and return the next
in line after "Derived" in the MRO: "Base".


How can I organize my code to make it easier to change the base class?
----------------------------------------------------------------------

You could assign the base class to an alias and derive from the alias.
Then all you have to change is the value assigned to the alias.
Incidentally, this trick is also handy if you want to decide
dynamically (e.g. depending on availability of resources) which base
class to use.  Example:

   class Base:
       ...

   BaseAlias = Base

   class Derived(BaseAlias):
       ...


How do I create static class data and static class methods?
-----------------------------------------------------------

Both static data and static methods (in the sense of C++ or Java) are
supported in Python.

For static data, simply define a class attribute.  To assign a new
value to the attribute, you have to explicitly use the class name in
the assignment:

   class C:
       count = 0   # number of times C.__init__ called

       def __init__(self):
           C.count = C.count + 1

       def getcount(self):
           return C.count  # or return self.count

"c.count" also refers to "C.count" for any "c" such that
"isinstance(c, C)" holds, unless overridden by "c" itself or by some
class on the base-class search path from "c.__class__" back to "C".

Caution: within a method of C, an assignment like "self.count = 42"
creates a new and unrelated instance named “count” in "self"’s own
dict.  Rebinding of a class-static data name must always specify the
class whether inside a method or not:

   C.count = 314

Static methods are possible:

   class C:
       @staticmethod
       def static(arg1, arg2, arg3):
           # No 'self' parameter!
           ...

However, a far more straightforward way to get the effect of a static
method is via a simple module-level function:

   def getcount():
       return C.count

If your code is structured so as to define one class (or tightly
related class hierarchy) per module, this supplies the desired
encapsulation.


How can I overload constructors (or methods) in Python?
-------------------------------------------------------

This answer actually applies to all methods, but the question usually
comes up first in the context of constructors.

In C++ you’d write

   class C {
       C() { cout << "No arguments\n"; }
       C(int i) { cout << "Argument is " << i << "\n"; }
   }

In Python you have to write a single constructor that catches all
cases using default arguments.  For example:

   class C:
       def __init__(self, i=None):
           if i is None:
               print("No arguments")
           else:
               print("Argument is", i)

This is not entirely equivalent, but close enough in practice.

You could also try a variable-length argument list, e.g.

   def __init__(self, *args):
       ...

The same approach works for all method definitions.


I try to use __spam and I get an error about _SomeClassName__spam.
------------------------------------------------------------------

Variable names with double leading underscores are “mangled” to
provide a simple but effective way to define class private variables.
Any identifier of the form "__spam" (at least two leading underscores,
at most one trailing underscore) is textually replaced with
"_classname__spam", where "classname" is the current class name with
any leading underscores stripped.

The identifier can be used unchanged within the class, but to access
it outside the class, the mangled name must be used:

   class A:
       def __one(self):
           return 1
       def two(self):
           return 2 * self.__one()

   class B(A):
       def three(self):
           return 3 * self._A__one()

   four = 4 * A()._A__one()

In particular, this does not guarantee privacy since an outside user
can still deliberately access the private attribute; many Python
programmers never bother to use private variable names at all.

See also:

  The private name mangling specifications for details and special
  cases.


My class defines __del__ but it is not called when I delete the object.
-----------------------------------------------------------------------

There are several possible reasons for this.

The "del" statement does not necessarily call "__del__()" – it simply
decrements the object’s reference count, and if this reaches zero
"__del__()" is called.

If your data structures contain circular links (e.g. a tree where each
child has a parent reference and each parent has a list of children)
the reference counts will never go back to zero.  Once in a while
Python runs an algorithm to detect such cycles, but the garbage
collector might run some time after the last reference to your data
structure vanishes, so your "__del__()" method may be called at an
inconvenient and random time. This is inconvenient if you’re trying to
reproduce a problem. Worse, the order in which object’s "__del__()"
methods are executed is arbitrary.  You can run "gc.collect()" to
force a collection, but there *are* pathological cases where objects
will never be collected.

Despite the cycle collector, it’s still a good idea to define an
explicit "close()" method on objects to be called whenever you’re done
with them.  The "close()" method can then remove attributes that refer
to subobjects.  Don’t call "__del__()" directly – "__del__()" should
call "close()" and "close()" should make sure that it can be called
more than once for the same object.

Another way to avoid cyclical references is to use the "weakref"
module, which allows you to point to objects without incrementing
their reference count. Tree data structures, for instance, should use
weak references for their parent and sibling references (if they need
them!).

Finally, if your "__del__()" method raises an exception, a warning
message is printed to "sys.stderr".


How do I get a list of all instances of a given class?
------------------------------------------------------

Python does not keep track of all instances of a class (or of a built-
in type). You can program the class’s constructor to keep track of all
instances by keeping a list of weak references to each instance.


Why does the result of "id()" appear to be not unique?
------------------------------------------------------

The "id()" builtin returns an integer that is guaranteed to be unique
during the lifetime of the object.  Since in CPython, this is the
object’s memory address, it happens frequently that after an object is
deleted from memory, the next freshly created object is allocated at
the same position in memory.  This is illustrated by this example:

>>> id(1000)
13901272
>>> id(2000)
13901272

The two ids belong to different integer objects that are created
before, and deleted immediately after execution of the "id()" call.
To be sure that objects whose id you want to examine are still alive,
create another reference to the object:

>>> a = 1000; b = 2000
>>> id(a)
13901272
>>> id(b)
13891296


When can I rely on identity tests with the *is* operator?
---------------------------------------------------------

The "is" operator tests for object identity.  The test "a is b" is
equivalent to "id(a) == id(b)".

The most important property of an identity test is that an object is
always identical to itself, "a is a" always returns "True".  Identity
tests are usually faster than equality tests.  And unlike equality
tests, identity tests are guaranteed to return a boolean "True" or
"False".

However, identity tests can *only* be substituted for equality tests
when object identity is assured.  Generally, there are three
circumstances where identity is guaranteed:

1. Assignments create new names but do not change object identity.
   After the assignment "new = old", it is guaranteed that "new is
   old".

2. Putting an object in a container that stores object references does
   not change object identity.  After the list assignment "s[0] = x",
   it is guaranteed that "s[0] is x".

3. If an object is a singleton, it means that only one instance of
   that object can exist.  After the assignments "a = None" and "b =
   None", it is guaranteed that "a is b" because "None" is a
   singleton.

In most other circumstances, identity tests are inadvisable and
equality tests are preferred.  In particular, identity tests should
not be used to check constants such as "int" and "str" which aren’t
guaranteed to be singletons:

   >>> a = 1000
   >>> b = 500
   >>> c = b + 500
   >>> a is c
   False

   >>> a = 'Python'
   >>> b = 'Py'
   >>> c = b + 'thon'
   >>> a is c
   False

Likewise, new instances of mutable containers are never identical:

   >>> a = []
   >>> b = []
   >>> a is b
   False

In the standard library code, you will see several common patterns for
correctly using identity tests:

1. As recommended by **PEP 8**, an identity test is the preferred way
   to check for "None".  This reads like plain English in code and
   avoids confusion with other objects that may have boolean values
   that evaluate to false.

2. Detecting optional arguments can be tricky when "None" is a valid
   input value.  In those situations, you can create a singleton
   sentinel object guaranteed to be distinct from other objects.  For
   example, here is how to implement a method that behaves like
   "dict.pop()":

      _sentinel = object()

      def pop(self, key, default=_sentinel):
          if key in self:
              value = self[key]
              del self[key]
              return value
          if default is _sentinel:
              raise KeyError(key)
          return default

3. Container implementations sometimes need to augment equality tests
   with identity tests.  This prevents the code from being confused by
   objects such as "float('NaN')" that are not equal to themselves.

For example, here is the implementation of
"collections.abc.Sequence.__contains__()":

   def __contains__(self, value):
       for v in self:
           if v is value or v == value:
               return True
       return False


How can a subclass control what data is stored in an immutable instance?
------------------------------------------------------------------------

When subclassing an immutable type, override the "__new__()" method
instead of the "__init__()" method.  The latter only runs *after* an
instance is created, which is too late to alter data in an immutable
instance.

All of these immutable classes have a different signature than their
parent class:

   from datetime import date

   class FirstOfMonthDate(date):
       "Always choose the first day of the month"
       def __new__(cls, year, month, day):
           return super().__new__(cls, year, month, 1)

   class NamedInt(int):
       "Allow text names for some numbers"
       xlat = {'zero': 0, 'one': 1, 'ten': 10}
       def __new__(cls, value):
           value = cls.xlat.get(value, value)
           return super().__new__(cls, value)

   class TitleStr(str):
       "Convert str to name suitable for a URL path"
       def __new__(cls, s):
           s = s.lower().replace(' ', '-')
           s = ''.join([c for c in s if c.isalnum() or c == '-'])
           return super().__new__(cls, s)

The classes can be used like this:

   >>> FirstOfMonthDate(2012, 2, 14)
   FirstOfMonthDate(2012, 2, 1)
   >>> NamedInt('ten')
   10
   >>> NamedInt(20)
   20
   >>> TitleStr('Blog: Why Python Rocks')
   'blog-why-python-rocks'


How do I cache method calls?
----------------------------

The two principal tools for caching methods are
"functools.cached_property()" and "functools.lru_cache()".  The former
stores results at the instance level and the latter at the class
level.

The *cached_property* approach only works with methods that do not
take any arguments.  It does not create a reference to the instance.
The cached method result will be kept only as long as the instance is
alive.

The advantage is that when an instance is no longer used, the cached
method result will be released right away.  The disadvantage is that
if instances accumulate, so too will the accumulated method results.
They can grow without bound.

The *lru_cache* approach works with methods that have *hashable*
arguments.  It creates a reference to the instance unless special
efforts are made to pass in weak references.

The advantage of the least recently used algorithm is that the cache
is bounded by the specified *maxsize*.  The disadvantage is that
instances are kept alive until they age out of the cache or until the
cache is cleared.

This example shows the various techniques:

   class Weather:
       "Lookup weather information on a government website"

       def __init__(self, station_id):
           self._station_id = station_id
           # The _station_id is private and immutable

       def current_temperature(self):
           "Latest hourly observation"
           # Do not cache this because old results
           # can be out of date.

       @cached_property
       def location(self):
           "Return the longitude/latitude coordinates of the station"
           # Result only depends on the station_id

       @lru_cache(maxsize=20)
       def historic_rainfall(self, date, units='mm'):
           "Rainfall on a given date"
           # Depends on the station_id, date, and units.

The above example assumes that the *station_id* never changes.  If the
relevant instance attributes are mutable, the *cached_property*
approach can’t be made to work because it cannot detect changes to the
attributes.

To make the *lru_cache* approach work when the *station_id* is
mutable, the class needs to define the "__eq__()" and "__hash__()"
methods so that the cache can detect relevant attribute updates:

   class Weather:
       "Example with a mutable station identifier"

       def __init__(self, station_id):
           self.station_id = station_id

       def change_station(self, station_id):
           self.station_id = station_id

       def __eq__(self, other):
           return self.station_id == other.station_id

       def __hash__(self):
           return hash(self.station_id)

       @lru_cache(maxsize=20)
       def historic_rainfall(self, date, units='cm'):
           'Rainfall on a given date'
           # Depends on the station_id, date, and units.


Modules
=======


How do I create a .pyc file?
----------------------------

When a module is imported for the first time (or when the source file
has changed since the current compiled file was created) a ".pyc" file
containing the compiled code should be created in a "__pycache__"
subdirectory of the directory containing the ".py" file.  The ".pyc"
file will have a filename that starts with the same name as the ".py"
file, and ends with ".pyc", with a middle component that depends on
the particular "python" binary that created it.  (See **PEP 3147** for
details.)

One reason that a ".pyc" file may not be created is a permissions
problem with the directory containing the source file, meaning that
the "__pycache__" subdirectory cannot be created. This can happen, for
example, if you develop as one user but run as another, such as if you
are testing with a web server.

Unless the "PYTHONDONTWRITEBYTECODE" environment variable is set,
creation of a .pyc file is automatic if you’re importing a module and
Python has the ability (permissions, free space, etc…) to create a
"__pycache__" subdirectory and write the compiled module to that
subdirectory.

Running Python on a top level script is not considered an import and
no ".pyc" will be created.  For example, if you have a top-level
module "foo.py" that imports another module "xyz.py", when you run
"foo" (by typing "python foo.py" as a shell command), a ".pyc" will be
created for "xyz" because "xyz" is imported, but no ".pyc" file will
be created for "foo" since "foo.py" isn’t being imported.

If you need to create a ".pyc" file for "foo" – that is, to create a
".pyc" file for a module that is not imported – you can, using the
"py_compile" and "compileall" modules.

The "py_compile" module can manually compile any module.  One way is
to use the "compile()" function in that module interactively:

   >>> import py_compile
   >>> py_compile.compile('foo.py')

This will write the ".pyc" to a "__pycache__" subdirectory in the same
location as "foo.py" (or you can override that with the optional
parameter "cfile").

You can also automatically compile all files in a directory or
directories using the "compileall" module.  You can do it from the
shell prompt by running "compileall.py" and providing the path of a
directory containing Python files to compile:

   python -m compileall .


How do I find the current module name?
--------------------------------------

A module can find out its own module name by looking at the predefined
global variable "__name__".  If this has the value "'__main__'", the
program is running as a script.  Many modules that are usually used by
importing them also provide a command-line interface or a self-test,
and only execute this code after checking "__name__":

   def main():
       print('Running test...')
       ...

   if __name__ == '__main__':
       main()


How can I have modules that mutually import each other?
-------------------------------------------------------

Suppose you have the following modules:

"foo.py":

   from bar import bar_var
   foo_var = 1

"bar.py":

   from foo import foo_var
   bar_var = 2

The problem is that the interpreter will perform the following steps:

* main imports "foo"

* Empty globals for "foo" are created

* "foo" is compiled and starts executing

* "foo" imports "bar"

* Empty globals for "bar" are created

* "bar" is compiled and starts executing

* "bar" imports "foo" (which is a no-op since there already is a
  module named "foo")

* The import mechanism tries to read "foo_var" from "foo" globals, to
  set "bar.foo_var = foo.foo_var"

The last step fails, because Python isn’t done with interpreting "foo"
yet and the global symbol dictionary for "foo" is still empty.

The same thing happens when you use "import foo", and then try to
access "foo.foo_var" in global code.

There are (at least) three possible workarounds for this problem.

Guido van Rossum recommends avoiding all uses of "from <module> import
...", and placing all code inside functions.  Initializations of
global variables and class variables should use constants or built-in
functions only.  This means everything from an imported module is
referenced as "<module>.<name>".

Jim Roskind suggests performing steps in the following order in each
module:

* exports (globals, functions, and classes that don’t need imported
  base classes)

* "import" statements

* active code (including globals that are initialized from imported
  values).

Van Rossum doesn’t like this approach much because the imports appear
in a strange place, but it does work.

Matthias Urlichs recommends restructuring your code so that the
recursive import is not necessary in the first place.

These solutions are not mutually exclusive.


__import__(‘x.y.z’) returns <module ‘x’>; how do I get z?
---------------------------------------------------------

Consider using the convenience function "import_module()" from
"importlib" instead:

   z = importlib.import_module('x.y.z')


When I edit an imported module and reimport it, the changes don’t show up.  Why does this happen?
-------------------------------------------------------------------------------------------------

For reasons of efficiency as well as consistency, Python only reads
the module file on the first time a module is imported.  If it didn’t,
in a program consisting of many modules where each one imports the
same basic module, the basic module would be parsed and re-parsed many
times.  To force re-reading of a changed module, do this:

   import importlib
   import modname
   importlib.reload(modname)

Warning: this technique is not 100% fool-proof.  In particular,
modules containing statements like

   from modname import some_objects

will continue to work with the old version of the imported objects.
If the module contains class definitions, existing class instances
will *not* be updated to use the new class definition.  This can
result in the following paradoxical behaviour:

   >>> import importlib
   >>> import cls
   >>> c = cls.C()                # Create an instance of C
   >>> importlib.reload(cls)
   <module 'cls' from 'cls.py'>
   >>> isinstance(c, cls.C)       # isinstance is false?!?
   False

The nature of the problem is made clear if you print out the
“identity” of the class objects:

   >>> hex(id(c.__class__))
   '0x7352a0'
   >>> hex(id(cls.C))
   '0x4198d0'
