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pythonregexscopegeneratoryield

python yield generator variable scope


I am using yield to create a generator that returns chunks of a string that are being extracted using a regex and re.sub(). While I found an approach that worked, I am a bit confused about why it works one way but not another, as shown below:

This doesn't work (processchunk() is not assigning to the chunk declared in splitmsg):

def splitmsg(msg):
    chunk = None
    def processchunk(match):
        chunk = match.group(1)
        return ""
    while True:
        chunk = None
        msg = re.sub(reCHUNK,processchunk,msg,1)
        if chunk:
            yield chunk
        else:
            break     

This does work (note the only difference being chunk is now a list chunks):

def splitmsg(msg):
    chunks = [ None, ]
    def processchunk(match):
        chunks[0] = match.group(1)
        return ""
    while True:
        chunks[0] = None
        msg = re.sub(reCHUNK,processchunk,msg,1)
        if chunks[0]:
            yield chunks[0]
        else:
            break

My question is basically why does it appear that the scoping of the chunk/chunks variable seem to depend on whether it is a plain variable or a list?


Solution

  • In python, variables can be 'pulled' from the surrounding scope if read from. So the following will work:

    def foo():
        spam = 'eggs'
        def bar():
            print spam
    foo()
    

    because the variable 'spam' is being looked up in the surrounding scope, the foo function.

    However, you cannot change the value of a surrounding scope. You can change global variables (if you declare them as global in your function), but you cannot do that for the variable spam in the above function.

    (Python 3 changes this, it adds a new keyword nonlocal. If you define spam as nonlocal inside of bar you can assign to that variable a new value inside of bar.)

    Now to your list. What happens there is that you are not altering the variable chunks at all. Throughout your code, chunks points to one list, and only to that list. As far as python is concerned, chunks the variable is not altered within the processchunk function.

    What does happen is that you alter the contents of the list. You can freely assign a new value to chunks[0], because that's not the variable chunks, it is the list referred to by chunks, first index. Python allows this because it is not a variable assignment, but a list manipulation instead.

    So, your 'workaround' is correct, if somewhat obscure. If you use Python 3, you can declare chunks as nonlocal within processchunk and then things will work without lists too.