Alternating between iterators in Python

Question

What is the most efficient way to alternate taking values from different iterators in Python, so that, for example, alternate(xrange(1, 7, 2), xrange(2, 8, 2)) would yield 1, 2, 3, 4, 5, 6. I know one way to implement it would be:

def alternate(*iters):
    while True:
        for i in iters:
            try:
                yield i.next()
            except StopIteration:
                pass

But is there a more efficient or cleaner way? (Or, better yet, an itertools function I missed?)

Answer 1

what about zip? you may also try izip from itertools

>>> zip(xrange(1, 7, 2),xrange(2, 8 , 2))
[(1, 2), (3, 4), (5, 6)]

if this is not what you want, please give more examples in your question post.

Answer 2

You could define alternate like this:

import itertools
def alternate(*iters):   
    for elt in itertools.chain.from_iterable(
        itertools.izip(*iters)):
        yield elt

print list(alternate(xrange(1, 7, 2), xrange(2, 8, 2)))

This leaves open the question of what to do if one iterator stops before another. If you'd like to continue until the longest iterator is exhausted, then you could use itertools.izip_longest in place of itertools.izip.

import itertools
def alternate(*iters):   
    for elt in itertools.chain.from_iterable(
        itertools.izip_longest(*iters)):
        yield elt
print list(alternate(xrange(1, 7, 2), xrange(2, 10, 2)))

This will put yield

[1, 2, 3, 4, 5, 6, None, 8]

Note None is yielded when the iterator xrange(1,7,2) raises StopIteration (has no more elements).

If you'd like to just skip the iterator instead of yielding None, you could do this:

Dummy=object()

def alternate(*iters):   
    for elt in itertools.chain.from_iterable(
        itertools.izip_longest(*iters,fillvalue=Dummy)):
        if elt is not Dummy:
            yield elt

Answer 3

For a "clean" implementation, you want

itertools.chain(*itertools.izip(*iters))

but maybe you want

itertools.chain(*itertools.izip_longest(*iters))

Answer 4

If they're the same length, itertools.izip can be leveraged like so:

def alternate(*iters):
    for row in itertools.izip(*iters):
       for i in row:
           yield i

Answer 5

See roundrobin in the itertools "Recipes" section. It's a more general version of alternate.