Question

The documentation for itertools provides a recipe for a pairwise() function, which I’ve slightly modified below so that it returns (last_item, None) as the final pair:

from itertools import tee, izip_longest

def pairwise_tee(iterable):
    a, b = tee(iterable)
    next(b, None)
    return izip_longest(a, b)

However, it seemed to me that using tee() might be overkill (given that it’s only being used to provide one step of look-ahead), so I tried writing an alternative that avoids it:

def pairwise_zed(iterator):
    a = next(iterator)
    for b in iterator:
        yield a, b
        a = b
    yield a, None

Note: it so happens that I know my input will be an iterator for my use case; I’m aware that the function above won’t work with a regular iterable. The requirement to accept an iterator is also why I’m not using something like izip_longest(iterable, iterable[1:]), by the way.

Testing both functions for speed gave the following results in Python 2.7.3:

>>> import random, string, timeit
>>> for length in range(0, 61, 10):
...     text = "".join(random.choice(string.ascii_letters) for n in range(length))
...     for variant in "tee", "zed":
...         test_case = "list(pairwise_%s(iter('%s')))" % (variant, text)
...         setup = "from __main__ import pairwise_%s" % variant
...         result = timeit.repeat(test_case, setup=setup, number=100000)
...         print "%2d %s %r" % (length, variant, result)
...     print
... 
 0 tee [0.4337780475616455, 0.42563915252685547, 0.42760396003723145]
 0 zed [0.21209311485290527, 0.21059393882751465, 0.21039700508117676]

10 tee [0.4933490753173828, 0.4958930015563965, 0.4938509464263916]
10 zed [0.32074403762817383, 0.32239794731140137, 0.32340312004089355]

20 tee [0.6139161586761475, 0.6109561920166016, 0.6153261661529541]
20 zed [0.49281787872314453, 0.49651598930358887, 0.4942781925201416]

30 tee [0.7470319271087646, 0.7446520328521729, 0.7463529109954834]
30 zed [0.7085139751434326, 0.7165200710296631, 0.7171430587768555]

40 tee [0.8083810806274414, 0.8031280040740967, 0.8049719333648682]
40 zed [0.8273730278015137, 0.8248250484466553, 0.8298079967498779]

50 tee [0.8745720386505127, 0.9205660820007324, 0.878741979598999]
50 zed [0.9760301113128662, 0.9776301383972168, 0.978381872177124]

60 tee [0.9913749694824219, 0.9922418594360352, 0.9938201904296875]
60 zed [1.1071209907531738, 1.1063809394836426, 1.1069209575653076]

… so, it turns out that pairwise_tee() starts to outperform pairwise_zed() when there are about forty items. That’s fine, as far as I’m concerned – on average, my input is likely to be under that threshold.

My question is: which should I use? pairwise_zed() looks like it’ll be a little faster (and to my eyes is slightly easier to follow), but pairwise_tee() could be considered the “canonical” implementation by virtue of being taken from the official docs (to which I could link in a comment), and will work for any iterable – which isn’t a consideration at this point, but I suppose could be later.

I was also wondering about potential gotchas if the iterator is interfered with outside the function, e.g.

for a, b in pairwise(iterator):
    # do something
    q = next(iterator)

… but as far as I can tell, pairwise_zed() and pairwise_tee() behave identically in that situation (and of course it would be a damn fool thing to do in the first place).

Answer 1

The itertools tee implementation is idiomatic for those experienced with itertools, though I’d be tempted to use islice instead of next to advance the leading iterator.

A disadvantage of your version is that it’s less easy to extend it to n-wise iteration as your state is stored in local variables; I’d be tempted to use a deque:

def pairwise_deque(iterator, n=2):
    it = chain(iterator, repeat(None, n - 1))
    d = collections.deque(islice(it, n - 1), maxlen=n)
    for a in it:
      d.append(a)
      yield tuple(d)

A useful idiom is calling iter on the iterator parameter; this is an easy way to ensure your function works on any iterable.