Python “Every Other Element” Idiom [duplicate]

Answer 1

toolz is a well-built library with many functional programming niceties overlooked in itertools. partition solves this (with an option to pad the last entry for lists of odd length)

>>> list(toolz.partition(2, [1,2,3,4,5,6]))
[(1, 2), (3, 4), (5, 6)]

Answer 2

This will do it a bit more neatly:

>>> data = [1,2,3,4,5,6]
>>> zip(data[0::2], data[1::2])
[(1, 2), (3, 4), (5, 6)]

(but it's arguably less readable if you're not familiar with the "stride" feature of ranges).

Like your code, it discards the last value where you have an odd number of values.

Answer 3

try this

def pairs(l, n):
    return zip(*[l[i::n] for i in range(n)])

So,

pairs([1, 2, 3, 4], 2) gives

[(1, 2), (3, 4)]

Answer 4

How about using the step feature of range():

[(l[n],l[n+1]) for n in range(0,len(l),2)]

Answer 5

I usually copy the grouper recipe from the itertools documentation into my code for this.

def grouper(n, iterable, fillvalue=None):
    "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

Answer 6

If you don't want to lose elements if their number in list is not even try this:

>>> l = [1, 2, 3, 4, 5]
>>> [(l[i],  l[i+1] if i+1 < len(l) else None)  for i in range(0, len(l), 2)]
[(1, 2), (3, 4), (5, None)]