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 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.
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)
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With