Python iterators – how to dynamically assign self.next within a new style class?

Question

As part of some WSGI middleware I want to write a python class that wraps an iterator to implement a close method on the iterator.

This works fine when I try it with an

Answer 1

Just return the iterator. That's what __iter__ is for. It makes no sense to try to monkey-patch the object into being in iterator and return it when you already have an iterator.

EDIT: Now with two methods. Once, monkey patching the wrapped iterator, second, kitty-wrapping the iterator.

class IteratorWrapperMonkey(object):

    def __init__(self, otheriter):
        self.otheriter = otheriter
        self.otheriter.close = self.close

    def close(self):
        print "Closed!"

    def __iter__(self):
        return self.otheriter

class IteratorWrapperKitten(object):

    def __init__(self, otheriter):
        self.otheriter = otheriter

    def __iter__(self):
        return self

    def next(self):
        return self.otheriter.next()

    def close(self):
        print "Closed!"

class PatchableIterator(object):

    def __init__(self, inp):
        self.iter = iter(inp)

    def next(self):
        return self.iter.next()

    def __iter__(self):
        return self

if __name__ == "__main__":
    monkey = IteratorWrapperMonkey(PatchableIterator([1, 2, 3]))
    for i in monkey:
        print i
    monkey.close()

    kitten = IteratorWrapperKitten(iter([1, 2, 3]))
    for i in kitten:
        print i
    kitten.close()

Both methods work both with new and old-style classes.