Why can't Python increment variable in closure

问题:

In this code snippet I can print the value of counter from inside the bar function

def foo():
    counter = 1
    def bar():
        print("bar", counter)
    return bar

bar = foo()

bar()

But when I try to increment counter from inside the bar function I get an UnboundLocalError error.

UnboundLocalError: local variable 'counter' referenced before assignment

Code snippet with increment statement in.

def foo():
    counter = 1
    def bar():
        counter += 1
        print("bar", counter)
    return bar

bar = foo()

bar()

Do you only have read access to variables in the outer function in a Python closure?

回答1:

You can't mutate closure variables in Python 2. In Python 3, which you appear to be using due to your print(), you can declare them nonlocal:

def foo():

  counter = 1

  def bar():
    nonlocal counter
    counter += 1
    print("bar", counter)

  return bar

Otherwise, the assignment within bar() makes the variable local, and since you haven't assigned a value to the variable in the local scope, trying to access it is an error.

In Python 2, my favorite workaround looks like this:

def foo():

  class nonlocal:
    counter = 1

  def bar():
    nonlocal.counter += 1
    print("bar", nonlocal.counter)

  return bar

This works because mutating a mutable object doesn't require changing what the variable name points to. In this case, nonlocal is the closure variable and it is never reassigned; only its contents are changed. Other workarounds use lists or dictionaries.

Or you could use a class for the whole thing, as @naomik suggests in a comment.

回答2:

Why can't Python increment variable in closure?

The error message is self-explanatory. I offer a couple of solutions here.

• Using a function attribute (uncommon, but works pretty well)
• Using a closure with nonlocal (ideal, but Python 3 only)
• Using a closure over a mutable object (idiomatic of Python 2)
• Using a method on a custom object
• Directly calling instance of the object by implementing __call__

Use an attribute on the function.

Set a counter attribute on your function manually after creating it:

def foo():
    foo.counter += 1
    return foo.counter

foo.counter = 0

And now:

>>> foo()
1
>>> foo()
2
>>> foo()
3

Or you can auto-set the function:

def foo():
    if not hasattr(foo, 'counter'):
        foo.counter = 0
    foo.counter += 1
    return foo.counter

Similarly:

>>> foo()
1
>>> foo()
2
>>> foo()
3

These approaches are simple, but uncommon, and unlikely to be quickly grokked by someone viewing your code without you present.

More common ways what you wish to accomplish is done varies depending on your version of Python.

Python 3, using a closure with nonlocal

In Python 3, you can declare nonlocal:

def foo():
    counter = 0
    def bar():
        nonlocal counter
        counter += 1
        print("bar", counter)
    return bar

bar = foo()

And it would increment

>>> bar()
bar 1
>>> bar()
bar 2
>>> bar()
bar 3

This is probably the most idiomatic solution for this problem. Too bad it's restricted to Python 3.

Python 2 workaround for nonlocal:

You could declare a global variable, and then increment on it, but that clutters the module namespace. So the idiomatic workaround to avoid declaring a global variable is to point to a mutable object that contains the integer on which you wish to increment, so that you're not attempting to reassign the variable name:

def foo():
    counter = [0]
    def bar():
        counter[0] += 1
        print("bar", counter)
    return bar

bar = foo()

and now:

>>> bar()
('bar', [1])
>>> bar()
('bar', [2])
>>> bar()
('bar', [3])

I do think that is superior to the suggestions that involve creating classes just to hold your incrementing variable. But to be complete, let's see that.

Using a custom object

class Foo(object):
    def __init__(self):
        self._foo_call_count = 0
    def foo(self):
        self._foo_call_count += 1
        print('Foo.foo', self._foo_call_count)

foo = Foo()

and now:

>>> foo.foo()
Foo.foo 1
>>> foo.foo()
Foo.foo 2
>>> foo.foo()
Foo.foo 3

or even implement __call__:

class Foo2(object):
    def __init__(self):
        self._foo_call_count = 0
    def __call__(self):
        self._foo_call_count += 1
        print('Foo', self._foo_call_count)

foo = Foo2()

and now:

>>> foo()
Foo 1
>>> foo()
Foo 2
>>> foo()
Foo 3

回答3:

As you can't modify immutable objects in Python, there is a workaround in Python 2.X:

Use List

def counter(start):
    count = [start]
    def incr():
        count[0] += 1
        return count[0]
    return incr

a = counter(100)
print a()
b = counter(200)
print b()

回答4:

nonlocal is a new statement in Python 3, and there is no equivalent in Python 2.

Useful link Understanding UnboundLocalError in Python