Try this:
l = [1, 2, 3, 4] for a, b in zip(*[iter(l)]*2): print a, b
zip(*[iter(seq)]*N) will group by N (but if there are any odd items at the end it will ignore them).
map(None, *[iter(seq)]*N) will group by N padding the last item with None if it needs to.
For anyone else who was as bemused as I was that Duncan's and F. Petitjean's suggestions actually *work*, this was what I had to do to figure out *why* they work:
Py> l = [1, 2, 3, 4] Py> itr = iter(l) Py> zip(itr) # Put all items from iterator in position 1 [(1,), (2,), (3,), (4,)] Py> itr = iter(l) Py> zip(itr, itr) # Put every second item in position 2 [(1, 2), (3, 4)]
Using zip(*[iter(l)]*N) or zip(*(iter(l),)*N) simply extends the above to the general case.
I'd definitely recommend hiding this trick inside a function. Perhaps something like (using Michael's function name):
from itertools import izip, repeat, chain
def partition(seq, part_len): return izip(*((iter(seq),) * part_len))
def padded_partition(seq, part_len, pad_val=None):
itr = iter(seq)
if (len(seq) % part_len != 0):
padding = repeat(pad_val, part_len)
itr = chain(itr, padding)
return izip(*((itr,) * part_len))Py> list(partition(range(10), 2)) [(0, 1), (2, 3), (4, 5), (6, 7), (8, 9)] Py> list(partition(range(10), 3)) [(0, 1, 2), (3, 4, 5), (6, 7, 8)] Py> list(padded_partition(range(10), 2)) [(0, 1), (2, 3), (4, 5), (6, 7), (8, 9)] Py> list(padded_partition(range(10), 3)) [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, None, None)] Py> list(padded_partition(range(10), 3, False)) [(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, False, False)] Py> zip(*padded_partition(range(10), 3)) [(0, 3, 6, 9), (1, 4, 7, None), (2, 5, 8, None)]
Not sure how useful that last example is, but I thought it was cute :)
Cheers, Nick.
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Nick Coghlan | [EMAIL PROTECTED] | Brisbane, Australia
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http://boredomandlaziness.skystorm.net
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