Python x if not None else y

It can be a surprise to get the 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

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

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

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

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 keyword:

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

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 for assigned variables provides a bar against unintended side-effects. On the other hand, if

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

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

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

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

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

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

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

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

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

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

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

This happens because

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

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

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

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

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

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

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

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

Here,

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

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

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

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

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

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

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

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

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

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

mod.py:

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

main.py:

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

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

In general, don’t use

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

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. , , ,

2. third-party library modules (anything installed in Python’s site-packages directory) – e.g.

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

   57,

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

   58,

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

   59

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 .

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

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

The first time you call this function,

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

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

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

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

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

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

Because of this feature, it is good programming practice to not use mutable objects as default values. Instead, use

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

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

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

but:

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

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:

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

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

Collect the arguments using the

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

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

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

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

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

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

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

foo, bar and kwargs are parameters of

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

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

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

the values

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

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

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

If you wrote code like:

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

you might be wondering why appending an element to

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

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

There are two factors that produce this result:

1. Variables are simply names that refer to objects. Doing

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

   79 doesn’t create a copy of the list – it creates a new variable

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

   77 that refers to the same object

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

   24 refers to. This means that there is only one object (the list), and both

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

   24 and

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

   77 refer to it.

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

After the call to

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

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

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

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

If we instead assign an immutable object to

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

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

we can see that in this case

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

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

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

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

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

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

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

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

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

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

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

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

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

Some operations (for example

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

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

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

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

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

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

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

However, there is one class of operations where the same operation sometimes has different behaviors with different types: the augmented assignment operators. For example,

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

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

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

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

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

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

In other words:

• If we have a mutable object (, , , 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 (, , , 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 operator, or the built-in function .

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:

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

   3

   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:

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

   4

4. By passing in a dictionary that gets mutated:

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

   5

5. Or bundle up values in a class instance:

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

   6

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

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

You have two choices: you can use nested scopes or you can use callable objects. For example, suppose you wanted to define

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

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

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

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

Or using a callable object:

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

In both cases,

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

gives a callable object where

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

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:

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

Object can encapsulate state for several methods:

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

Here

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

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

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

In general, try or for the general case. Not all objects can be copied, but most can.

Some objects can be copied more easily. Dictionaries have a method:

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

Sequences can be copied by slicing:

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

For an instance

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

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

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

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

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

Arguably the class has a name: even though it is bound to two names and invoked through the name

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

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

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

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

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!

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

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

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

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

not:

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

The same is true of the various assignment operators (

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

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

Yes, there is. The syntax is as follows:

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

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

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

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

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

Yes. Usually this is done by nesting within

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

>>> print(x)
11

Don’t try this at home, kids!

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, is a function that accepts positional-only parameters. Its documentation looks like this:

>>> print(x)
11

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

>>> print(x)
11

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:

>>> print(x)
11

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:

>>> print(x)
11

It’s primarily driven by the desire that

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

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

>>> print(x)
11

then integer division has to return the floor. C also requires that identity to hold, and then compilers that truncate

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

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

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

There are few real use cases for

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

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

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

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

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

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

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

Trying to lookup an

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

>>> print(x)
11

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

>>> print(x)
11

For integers, use the built-in type constructor, e.g.

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

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

By default, these interpret the number as decimal, so that

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

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

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

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

Do not use the built-in function if all you need is to convert strings to numbers. 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

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

also has the effect of interpreting numbers as Python expressions, so that e.g.

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

To convert, e.g., the number

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

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

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

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

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

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

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 object or the module:

>>> print(x)
11

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:

  >>> print(x)
  11
  

  9

• Use the built-in function :

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

  0

  Note that works on any object, including classes, class instances, modules, and so on.

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

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

  1

• Use to resolve the function name:

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

  2

You can use

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

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

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

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

Since this is typically only desired when reading text one line at a time, using

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

Not as such.

For simple input parsing, the easiest approach is usually to split the line into whitespace-delimited words using the method of string objects and then convert decimal strings to numeric values using or .

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

For more complicated input parsing, regular expressions are more powerful than C’s

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

See the .

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

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

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

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

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

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

It is also possible to use to append a backslash on Windows:

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

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:

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

Also see the specification in the .

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 .

• 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 module).

• Writing benchmark scripts will allow you to iterate quickly when searching for improvements (see the 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 and the 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 built-in method or the related function to do sorting (and see the 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 yourself.

See also

The wiki page devoted to performance tips.

and 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 objects, the recommended idiom is to place them into a list and call at the end:

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

(another reasonably efficient idiom is to use )

To accumulate many objects, the recommended idiom is to extend a object using in-place concatenation (the

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

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

The type constructor

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

For example,

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

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

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

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

The type constructor

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

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

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

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

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

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

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

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

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

Using negative indices can be very convenient. For example

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

Use the built-in function:

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

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

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:

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

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

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

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

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

The list comprehension may be fastest.

Use a list:

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

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

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

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

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

If mutability is desired, you could use lists instead of tuples. Here the analogue of a Lisp car is

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

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

You probably tried to make a multidimensional array like this:

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

This looks correct if you print it:

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

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

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

The reason is that replicating a list with

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

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

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

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

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

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

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

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

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

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

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

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

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

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

If you wrote:

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

The reason for the exception should be immediately clear:

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

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

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

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

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

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

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

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

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

When you write something like:

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

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

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

To see why this happens, you need to know that (a) if an object implements an magic method, it gets called when the

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

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

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

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

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

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

This is equivalent to:

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

The object pointed to by a_list has been mutated, and the pointer to the mutated object is assigned back to

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

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

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

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

The

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

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

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

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

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

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

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

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

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

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

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

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

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

A method is a function on some object

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

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

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

Self is merely a conventional name for the first argument of a method. A method defined as

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

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

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

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

See also .

Use the built-in function . 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.

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

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

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

Note that also checks for virtual inheritance from an . So, the test will return

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

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

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

Note that most programs do not use 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:

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

A better approach is to define a

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

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

Delegation is an object oriented technique (also called a design pattern). Let’s say you have an object

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

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

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:

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

Here the

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

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

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

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

Note that for more general cases delegation can get trickier. When attributes must be set as well as retrieved, the class must define a method too, and it must do so carefully. The basic implementation of

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

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

Most

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

Use the built-in function:

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

In the example, will automatically determine the instance from which it was called (the

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

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

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

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

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:

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

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:

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

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

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

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

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

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

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

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

Caution: within a method of C, an assignment like

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

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

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

Static methods are possible:

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

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

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

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

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

In C++ you’d write

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

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

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

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

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

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

The same approach works for all method definitions.

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

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

This doesn’t guarantee privacy: an outside user can still deliberately access the “_classname__spam” attribute, and private values are visible in the object’s

>>> print(x)
11

There are several possible reasons for this.

The statement does not necessarily call – it simply decrements the object’s reference count, and if this reaches zero

>>> print(x)
11

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

>>> print(x)
11

>>> print(x)
11

Despite the cycle collector, it’s still a good idea to define an explicit

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

Another way to avoid cyclical references is to use the 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

>>> print(x)
11

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.

The 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:

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

The two ids belong to different integer objects that are created before, and deleted immediately after execution of the

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

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

The

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

>>> print(x)
11

>>> print(x)
11

The most important property of an identity test is that an object is always identical to itself,

>>> print(x)
11

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

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

>>> print(x)
11

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

>>> print(x)
11

>>> print(x)
11

2) Putting an object in a container that stores object references does not change object identity. After the list assignment

>>> print(x)
11

>>> print(x)
11

3) If an object is a singleton, it means that only one instance of that object can exist. After the assignments

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

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

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 and which aren’t guaranteed to be singletons:

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

Likewise, new instances of mutable containers are never identical:

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

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

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

2) Detecting optional arguments can be tricky when

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

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

3) Container implementations sometimes need to augment equality tests with identity tests. This prevents the code from being confused by objects such as

>>> print(x)
11

For example, here is the implementation of

>>> print(x)
11

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

When subclassing an immutable type, override the method instead of the 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:

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

The classes can be used like this:

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

The two principal tools for caching methods are and . 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:

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

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 and methods so that the cache can detect relevant attribute updates:

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

When a module is imported for the first time (or when the source file has changed since the current compiled file was created) a

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

One reason that a

>>> print(x)
11

>>> print(x)
11

Unless the 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

>>> print(x)
11

Running Python on a top level script is not considered an import and no

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

If you need to create a

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

The module can manually compile any module. One way is to use the

>>> print(x)
11

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

This will write the

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

You can also automatically compile all files in a directory or directories using the module. You can do it from the shell prompt by running

>>> print(x)
11

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

A module can find out its own module name by looking at the predefined global variable

>>> print(x)
11

>>> print(x)
11

>>> print(x)
11

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

Suppose you have the following modules:

>>> print(x)
11

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

>>> print(x)
11

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

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

• main imports

  >>> print(x)
  11
  

  65

• Empty globals for

  >>> print(x)
  11
  

  65 are created

• >>> print(x)
  11
  

  65 is compiled and starts executing

• >>> print(x)
  11
  

  65 imports

  >>> print(x)
  11
  

  95

• Empty globals for

  >>> print(x)
  11
  

  95 are created

• >>> print(x)
  11
  

  95 is compiled and starts executing

• >>> print(x)
  11
  

  95 imports

  >>> print(x)
  11
  

  65 (which is a no-op since there already is a module named

  >>> print(x)
  11
  

  65)

• The import mechanism tries to read

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

  01 from

  >>> print(x)
  11
  

  65 globals, to set

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

  03

The last step fails, because Python isn’t done with interpreting

>>> print(x)
11

>>> print(x)
11

The same thing happens when you use

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

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

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

Guido van Rossum recommends avoiding all uses of

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

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

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

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

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

  10 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.

Consider using the convenience function from instead:

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

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:

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

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

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

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:

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