Python Programming/Scoping

Variables

Variables in Python are automatically declared by assignment. Variables are always references to objects, and are never typed. Variables exist only in the current scope or global scope. When they go out of scope, the variables are destroyed, but the objects to which they refer are not (unless the number of references to the object drops to zero).

Scope is delineated by function and class blocks. Both functions and their scopes can be nested. So therefore

def foo():
    def bar():
        x = 5 # x is now in scope
        return x + y # y is defined in the enclosing scope later
    y = 10
    return bar() # now that y is defined, bar's scope includes y

Now when this code is tested,

>>> foo()
15
>>> bar()
Traceback (most recent call last):
  File "<pyshell#26>", line 1, in -toplevel-
    bar()
NameError: name 'bar' is not defined

The name 'bar' is not found because a higher scope does not have access to the names lower in the hierarchy.

It is a common pitfall to fail to assign an object to a variable before use. In its most common form:

>>> for x in range(10):
         y.append(x) # append is an attribute of lists

Traceback (most recent call last):
  File "<pyshell#46>", line 2, in -toplevel-
    y.append(x)
NameError: name 'y' is not defined

Here, to correct this problem, one must add y = [] before the for loop executes.

A loop does not create its own scope:

for x in [1, 2, 3]:
  inner = x
print(inner) # 3 rather than an error

Keyword global

Global variables of a Python module are read-accessible from functions in that module. In fact, if they are mutable, they can be also modified via method call. However, they cannot be modified by a plain assignment unless they are declared global in the function.

An example to clarify:

count1 = 1
count2 = 1
list1 = []
list2 = []

def test1():
  print(count1)   # Read access is unproblematic, referring to the global

def test2():
  try:
    print(count1) # This try block is problematic because...
    count1 += 1   # count1 += 1 causes count1 to be local but local version is undefined.
  except UnboundLocalError as error:
    print("Error caught:", error)

def test3():
  list1 = [2]     # No outside effect; this defines list1 to be a local variable

def test4():
  global count2, list2
  print(count1)   # Read access is unproblematic, referring to the global
  count2 += 1     # We can modify count2 via assignment
  list1.append(1) # Impacts the global list1 even without global declaration since its a method call
  list2 = [2]     # We can modify list2 via assignment 

test1()
test2()
test3()
test4()

print("count1:", count1) # 1
print("count2:", count2) # 2
print("list1:", list1)   # [1]
print("list2:", list2)   # [2]

Links:

• 6.13. The global statement, docs.python.org
• What are the rules for local and global variables in Python? in Programming FAQ, docs.python.org

Keyword nonlocal

Keyword nonlocal, available since Python 3.0, is an analogue of global for nested scopes. It enables a nested function to assign-modify even an immutable variable that is local to the outer function.

An example:

# Requires Python 3
def outer():
  outerint = 0
  outerint2 = 10
  def inner():
    nonlocal outerint
    outerint = 1 # Impacts outer's outerint only because of the nonlocal declaration
    outerint2 = 1 # No impact
  inner()
  print(outerint)
  print(outerint2)

outer()

Simulation of nonlocal in Python 2 via a mutable object:

def outer():
  outerint = [1]           # Technique 1: Store int in a list
  class outerNL: pass      # Technique 2: Store int in a class
  outerNL.outerint2 = 11
  def inner():
    outerint[0] = 2        # List members can be modified
    outerNL.outerint2 = 12 # Class members can be modified
  inner()
  print(outerint[0])
  print(outerNL.outerint2)

outer()

Links:

• 7.13. The nonlocal statement, docs.python.org

globals and locals

To find out which variables exist in the global and local scopes, you can use locals() and globals() functions, which return dictionaries:

int1 = 1
def test1():
  int1 = 2
  globals()["int1"] = 3  # Write access seems possible
  print(locals()["int1"])# 2
  
test1()

print(int1)              # 3

Write access to locals() dictionary is discouraged by the Python documentation.

Links:

• 2. Built-in Functions # globals, docs.python.org
• 2. Built-in Functions # locals, docs.python.org

External links

• 4. Execution model, docs.python.org
• 7.13. The nonlocal statement, docs.python.org
• PEP 3104 -- Access to Names in Outer Scopes, python.org