Return from an iterator and then throw StopIteration

Question

What would be the nice way to return something from an iterator one last time when it's exhausted. I'm using a flag, but this is rather ugly:

class Example():

    def __iter__(self):
        self.lst = [1,2,3]
        self.stop = False # <-- ugly            
        return self

    def next(self):
        if self.stop:  # <-- ugly
            raise StopIteration
        if len(self.lst) == 0:
            self.stop = True            
            return "one last time"
        return self.lst.pop()

Background: I'm fetching an unknown amount of strings from an external source and send them further down to the caller. When the process is over, I want to emit a string "x records processed". I have no control over calling code, so this must be done inside my iterator.

score 6 · Answer 1 · answered Feb 22 '12 at 18:53

6

Maybe you can use a generator function instead:

def example():
    lst = [1, 2, 3]
    while lst:
        yield lst.pop()
    yield 'one last time'

answered Feb 22 '12 at 18:53

Dan Gerhardsson

1,869
13
12

3

Keep in mind that you can make an object's `__iter__` method be a generator function itself, in cases where you can't or don't want to use a generator directly. – Thomas Wouters Feb 22 '12 at 19:01
Thanks, "yield" is what I missed. – georg Feb 22 '12 at 19:18

Zach Kelling · Accepted Answer · 2012-02-22T19:38:43.630

6

You could just yield from __iter__ which would turn it into a generator function (alternately you could just write a generator function as suggested by Dan). Just as a warning, this might be misleading to people that abuse the next method.

class Example():

    def __iter__(self):
        lst = [1,2,3]
        for i in reversed(lst):
            yield i
        yield "one last time"

edited Feb 22 '12 at 19:38

answered Feb 22 '12 at 18:59

Zach Kelling

52,505
13
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108

score 1 · Answer 3 · answered Feb 22 '12 at 19:19

1

Don't return one extra thing. That's a bad idea because it doesn't extend well. What if you want to do sum as well as count? Or hash as well as count? An iterator is a stateful object. Make use of that.

class Example( collections.Iterator ):
    def __iter__(self):
        self.lst = [1,2,3]
        self.count = 0       
        return self
    def next(self):
        if self.lst:
            self.count += 1
            return self.lst.pop()
        raise StopIteration()

Use it like this.

my_iter= iter(Example())
for item in my_iterprin:
    print( item )

print( my_iter.count )

answered Feb 22 '12 at 19:19

S.Lott

384,516
81
508
779

1

I can't do that because I don't control the calling code (it's a wsgi app). – georg Feb 22 '12 at 19:32
1

I don't control the calling code? That's rarely true. Call them on the phone and talk to them. – S.Lott Feb 22 '12 at 19:43

score 0 · Answer 4 · answered Feb 22 '12 at 18:55

You could do something like this:

class Example():
    def __iter__(self):
        self.lst = [1, 2, 3]
        return self

    def next(self):
        try:
            return self.lst.pop()
        except IndexError:
            print "done iterating"
            raise StopIteration

>>> for i in Example():
...   print i
...
3
2
1
done iterating

In your actual code you will probably need to change the exception type that you are catching, but this format should still be applicable.

Return from an iterator and then throw StopIteration

4 Answers4