[Tutor] Process list elements as consecutive pairs

[Hugo Arts]

On Fri, Mar 5, 2010 at 10:48 PM, Hugo Arts <hugo.yoshi at gmail.com> wrote:
>
> Except the OP requested pairs (1, 2), (3, 4), i.e. with no duplicate
> elements. Here is a generator that does what you need:
>
> def pairs(seq):
>    it = iter(seq)
>    try:
>        while True:
>            yield it.next(), it.next()
>    except StopIteration:
>        return
>
> Hugo
>

Just Noticed this tiny caveat: If the length of the sequence is
uneven, the last element will not be yielded by this generator.
Whether that's a problem, and if it is, how to handle it, depends on
the application you're working with.

I did make a tiny modification to yield the last element in a pair
with None as the second value. The benefit is that it doesn't skip any
values, but you'll need to handle the possible None value in your
code. Of course, if you want a generator that yields consecutive
pairs, passing it a sequence of uneven length is sort of problematic
to begin with. So for most applications the simple version should be
fine.

def pairs(seq):
    it = iter(seq)
    try:
        while True:
            a = it.next()
            b = it.next()
            yield a, b
    except StopIteration:
        if len(seq) % 2:
            yield a, None

It's not as pretty as the simple version, unfortunately. in my
experience, corner cases are rarely handleable by elegant code :(

Compare output of the first and second version:

>>> # first (elegant) version of pairs
>>> list(pairs(range(5)))
[(0, 1), (2, 3)]
>>> # second version
>>> list(pairs_2(range(5)))
[(0, 1), (2, 3), (4, None)]