Need help understanding Python closures

Possible Duplicate:
UnboundLocalError in Python

I have this code:

import re

def doReplace(toReplace):
    i = 1
    def chapterReplacer(_):
        result = 'Chapter %i' % i
        i += 1
        return result

    return re.sub('Chapter [a-zA-Z]+', chapterReplacer, test)

test = 'Chapter one Chapter Two Chapter three'
print doReplace(test)

when I run it, I get the following error:

Traceback (most recent call last):
  File "C:/Python26/replace.py", line 13, in <module>
    print doReplace(test)
  File "C:/Python26/replace.py", line 10, in doReplace
    return re.sub('Chapter [a-zA-Z]+', chapterReplacer, test)
  File "C:Python26libre.py", line 151, in sub
    return _compile(pattern, 0).sub(repl, string, count)
  File "C:/Python26/replace.py", line 6, in chapterReplacer
    result = 'Chapter %i' % i
UnboundLocalError: local variable 'i' referenced before assignment

I was under the impression that chapterReplacer would capture the local variable i, but that doesn't seem to be happening?

---

You can make i a function attribute

def doReplace(toReplace):
    chapterReplacer.i = 1
    def chapterReplacer(_):
        result = 'Chapter %i' % chapterReplacer.i
        chapterReplacer.i += 1
        return result

    return re.sub('Chapter [a-zA-Z]+', chapterReplacer, test)

EDIT: As of python 3, you can use nonlocal a la @MartijnPieters 's solution.

---

Nope, and in python 2 you can't at all without resorting to using tricks with mutables:

def doReplace(toReplace):
    i = [1]
    def chapterReplacer(_):
        result = 'Chapter %i' % i[0]
        i[0] += 1
        return result

    return re.sub('Chapter [a-zA-Z]+', chapterReplacer, test)

Normally, python will only look in the surrounding scope for a variable if it is not being assigned to locally; as soon as the bytecompiler sees a direct assignment ( i = something ) and no global i statement to persuade it otherwise, a variable is considered local.

But in the above code we never assign to i in the chapterReplacer function. Yes, we do change i[0] but the value stored in i itself, a list, does not change.

In python 3, just use the nonlocal statement to have python look in it's closure for the variable:

def doReplace(toReplace):
    i = 1
    def chapterReplacer(_):
        nonlocal i
        result = 'Chapter %i' % i
        i += 1
        return result

    return re.sub('Chapter [a-zA-Z]+', chapterReplacer, test)

---

在Python中，如果您在函数内部赋值变量（即使使用复合赋值运算符（例如+= ）），该变量将被视为本地变量，除非global或非nonlocal语句另有指定。

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