views:

68

answers:

1

Given a list

A = [1 2 3 4 5 6]

Is there any idiomatic (Pythonic) way to iterate over it as though it were

B = [(1, 2) (3, 4) (5, 6)]

other than indexing? That feels like a holdover from C:

for a1,a2 in [ (A[i], A[i+1]) for i in range(0, len(A), 2) ]:

I can't help but feel there should be some clever hack using itertools or slicing or something.

(Of course, two at a time is just an example; I'd like a solution that works for any n.)

Edit: related http://stackoverflow.com/questions/1162592/iterate-over-a-string-2-or-n-characters-at-a-time-in-python but even the cleanest solution (accepted, using zip) doesn't generalize well to higher n without a list comprehension and *-notation.

+4  A: 

From http://docs.python.org/library/itertools.html:

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

i = grouper(3,range(100))
i.next()
(0, 1, 2)
MattH
This may not be subjectively idiomatic or avoid list comprehensions and * notation, but it is "pythonic" enough to be in the documentation.
MattH
Ah, I knew it existed somewhere. Thank you!
Wang