Short description of the scoping rules?

Question

What exactly are the Python scoping rules?

If I have some code:

code1
class Foo:
   code2
   def spam.....
      code3
      for code4..:
       code5
       x()

Where is x found? Some possible choices include the list below:

1. In the enclosing source file
2. In the class namespace
3. In the function definition
4. In the for loop index variable
5. Inside the for loop

Also there is the context during execution, when the function spam is passed somewhere else. And maybe lambda functions pass a bit differently?

There must be a simple reference or algorithm somewhere. It's a confusing world for intermediate Python programmers.

Answer 1

Python resolves your variables with -- generally -- three namespaces available.

At any time during execution, there are at least three nested scopes whose namespaces are directly accessible: the innermost scope, which is searched first, contains the local names; the namespaces of any enclosing functions, which are searched starting with the nearest enclosing scope; the middle scope, searched next, contains the current module's global names; and the outermost scope (searched last) is the namespace containing built-in names.

There are two functions: globals and locals which show you the contents two of these namespaces.

Namespaces are created by packages, modules, classes, object construction and functions. There aren't any other flavors of namespaces.

In this case, the call to a function named x has to be resolved in the local name space or the global namespace.

Local in this case, is the body of the method function Foo.spam.

Global is -- well -- global.

The rule is to search the nested local spaces created by method functions (and nested function definitions), then search global. That's it.

There are no other scopes. The for statement (and other compound statements like if and try) don't create new nested scopes. Only definitions (packages, modules, functions, classes and object instances.)

Inside a class definition, the names are part of the class namespace. code2, for instance, must be qualified by the class name. Generally Foo.code2. However, self.code2 will also work because Python objects look at the containing class as a fall-back.

An object (an instance of a class) has instance variables. These names are in the object's namespace. They must be qualified by the object. (variable.instance.)

From within a class method, you have locals and globals. You say self.variable to pick the instance as the namespace. You'll note that self is an argument to every class member function, making it part of the local namespace.

See Python Scope Rules, Python Scope, Variable Scope.

Answer 2

Essentially, the only thing in Python that introduces a new scope is a function definition. Classes are a bit of a special case in that anything defined directly in the body is placed in the class's namespace, but they are not directly accessible from within the methods (or nested classes) they contain.

In your example there are only 3 scopes where x will be searched in:

• spam's scope - containing everything defined in code3 and code5 (as well as code4, your loop variable)

• The global scope - containing everything defined in code1, as well as Foo (and whatever changes after it)

• The builtins namespace. A bit of a special case - this contains the various Python builtin functions and types such as len() and str(). Generally this shouldn't be modified by any user code, so expect it to contain the standard functions and nothing else.

More scopes only appear when you introduce a nested function (or lambda) into the picture. These will behave pretty much as you'd expect however. The nested function can access everything in the local scope, as well as anything in the enclosing function's scope. eg.

def foo():
    x=4
    def bar():
        print x  # Accesses x from foo's scope
    bar()  # Prints 4
    x=5
    bar()  # Prints 5

Restrictions:

Variables in scopes other than the local function's variables can be accessed, but can't be rebound to new parameters without further syntax. Instead, assignment will create a new local variable instead of affecting the variable in the parent scope. For example:

global_var1 = []
global_var2 = 1

def func():
    # This is OK: It's just accessing, not rebinding
    global_var1.append(4) 

    # This won't affect global_var2. Instead it creates a new variable
    global_var2 = 2 

    local1 = 4
    def embedded_func():
        # Again, this doen't affect func's local1 variable.  It creates a 
        # new local variable also called local1 instead.
        local1 = 5
        print local1

    embedded_func() # Prints 5
    print local1    # Prints 4

In order to actually modify the bindings of global variables from within a function scope, you need to specify that the variable is global with the global keyword. Eg:

global_var = 4
def change_global():
    global global_var
    global_var = global_var + 1

Currently there is no way to do the same for variables in enclosing function scopes, but Python 3 introduces a new keyword, "nonlocal" which will act in a similar way to global, but for nested function scopes.

Answer 3

Where is x found?

x is not found as you haven't defined it. :-) It could be found in code1 (global) or code3 (local) if you put it there.

code2 (class members) aren't visible to code inside methods of the same class — you would usually access them using self. code4/code5 (loops) live in the same scope as code3, so if you wrote to x in there you would be changing the x instance defined in code3, not making a new x.

Python is statically scoped, so if you pass ‘spam’ to another function spam will still have access to globals in the module it came from (defined in code1), and any other containing scopes (see below). code2 members would again be accessed through self.

lambda is no different to def. If you have a lambda used inside a function, it's the same as defining a nested function. In Python 2.2 onwards, nested scopes are available. In this case you can bind x at any level of function nesting and Python will pick up the innermost instance:

x= 0
def fun1():
    x= 1
    def fun2():
        x= 2
        def fun3():
            return x
        return fun3()
    return fun2()
print fun1(), x

2 0

fun3 sees the instance x from the nearest containing scope, which is the function scope associated with fun2. But the other x instances, defined in fun1 and globally, are not affected.

Before nested_scopes — in Python pre-2.1, and in 2.1 unless you specifically ask for the feature using a from-future-import — fun1 and fun2's scopes are not visible to fun3, so S.Lott's answer holds and you would get the global x:

0 0

Answer 4

The scoping rules for Python 2.x have been outlined already in other answers. The only thing I would add is that in Python 3.0, there is also the concept of a non-local scope (indicated by the 'nonlocal' keyword). This allows you to access outer scopes directly, and opens up the ability to do some neat tricks, including lexical closures (without ugly hacks involving mutable objects).

EDIT: Here's the PEP with more information on this.

Answer 5

Actually, a concise rule for Python Scope resolution, from Learning Python, 3rd. Ed.. (These rules are specific to variable names, not attributes. If you reference it without a period, these rules apply.)

LEGB Rule

• Local — Names assigned in any way within a function (def or lambda), and not declared global in that function

• Enclosing-function — Names assigned in the local scope of any and all statically enclosing functions (def or lambda), from inner to outer

• Global (module) — Names assigned at the top-level of a module file, or by executing a global statement in a def within the file

• Built-in (Python) — Names preassigned in the built-in names module: open, range, SyntaxError, etc

So, in the case of

code1
class Foo:
    code2
    def spam():
        code3
        for code4:
            code5
            x()

The for loop does not have its own namespace. In LEGB order, the scopes would be

• L: Local in def spam (in code3, code4, and code5)
• E: Any enclosing functions (if the whole example were in another def)
• G: Were there any x declared globally in the module (in code1)?
• B: Any builtin x in Python.

x will never be found in code2 (even in cases where you might expect it would, see Antti's answer or here).

Answer 6

In Python,

any variable that is assigned a value is local to the block in which the assignment appears.

If a variable can't be found in the current scope, please refer to the LEGB order.