Alternating between iterators in Python [duplicate]

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?)

Very similar question: stackoverflow.com/questions/243865/… — Seth
– Seth, Commented Jan 7, 2010 at 2:55
@Sean Devlin: No, 1 2 3 4 5 6 is correct. docs.python.org/library/functions.html#xrange — BlueRaja - Danny Pflughoeft
– BlueRaja - Danny Pflughoeft, Commented Jan 7, 2010 at 3:13

user245231 · Accepted Answer · 2010-01-07 02:55:35Z

18

For a "clean" implementation, you want

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

but maybe you want

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

answered Jan 7, 2010 at 2:55

user245231

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2 Comments

Dan Over a year ago

Does this exhaust the generator before passing it on to izip and then chain?

Abbafei Over a year ago

@Dan If iters was a generator, it would exhaust it, at least in python 2; running def z(*iters): print(type(iters)); print(iters) then z(*(i for i in xrange(1,18))) prints <type 'tuple'> and (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17).

ghostdog74 · Accepted Answer · 2010-01-07 02:51:15Z

8

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.

answered Jan 7, 2010 at 2:51

ghostdog74

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3 Comments

LeafStorm Over a year ago

On its own, izip isn't enough for my purposes. But putting a chain around izip, like chain.from_iterable(izip(iterables)) does work. I guess that's the cool thing about iterables. Thanks!

Sapph Over a year ago

The trouble with zip in this instance is that it will evaluate the iterators immediately, forcing you to forego generator semantics.

LeafStorm Over a year ago

Which would be especially problematic considering that my original use case for this was iterators that generated an infinite number of values.

Matthieu M. · Accepted Answer · 2014-12-18 14:30:48Z

7

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

def roundrobin(*iterables):
    "roundrobin('ABC', 'D', 'EF') --> A D E B F C"
    # Recipe credited to George Sakkis
    pending = len(iterables)
    nexts = cycle(iter(it).__next__ for it in iterables)
    while pending:
        try:
            for next in nexts:
                yield next()
        except StopIteration:
            pending -= 1
            nexts = cycle(islice(nexts, pending))

edited Dec 18, 2014 at 14:30

Matthieu M.

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answered Jan 7, 2010 at 10:04

David Jones

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Comments

Ethan Furman · Accepted Answer · 2016-03-23 05:02:54Z

3

There are two problems with your attempt:

You don't wrap each object in iters with iter() so it will fail with iterables such as list; and
By passing on StopIteration your generator is an infinite loop.

Some simple code that does solves both those issues and is still easy to read and understand:

def alternate(*iters):
    iters = [iter(i) for i in iters]
    while True:
        for i in iters:
            yield next(i)

>>> list(alternate(range(1, 7, 2), range(2, 8, 2)))
[1, 2, 3, 4, 5, 6]

edited Mar 23, 2016 at 5:02

answered Mar 23, 2016 at 2:42

Ethan Furman

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Comments

Crast · Accepted Answer · 2010-01-07 02:55:44Z

2

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

answered Jan 7, 2010 at 2:55

Crast

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Comments

unutbu · Accepted Answer · 2010-01-07 04:05:26Z

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

edited Jan 7, 2010 at 4:05

answered Jan 7, 2010 at 2:54

unutbu

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1 Comment

Chris Lutz Over a year ago

Might be cleaner to define Dummy as Dummy = object()

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