itertools question

[Neal Becker]

Is there any canned iterator adaptor that will

transform:
in = [1,2,3....]

into:
out = [(1,2,3,4), (5,6,7,8),...]

That is, each time next() is called, a tuple of the next N items is
returned.

 

[Glauco]

Neal Becker ha scritto:

Is there any canned iterator adaptor that will

transform:
in = [1,2,3....]

into:
out = [(1,2,3,4), (5,6,7,8),...]

That is, each time next() is called, a tuple of the next N items is
returned.

http://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-sized-chunks-in-python



Gla

 

[Peter Otten]

Is there any canned iterator adaptor that will

transform:
in = [1,2,3....]

into:
out = [(1,2,3,4), (5,6,7,8),...]

That is, each time next() is called, a tuple of the next N items is
returned.

Depending on what you want to do with items that don't make a complete N-
tuple:
[(0, 1, 2), (3, 4, 5), (6, 7, 8)]
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, None, None)]
3))))))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9,)]

Peter

 

[pataphor]

Is there any canned iterator adaptor that will

transform:
in = [1,2,3....]

into:
out = [(1,2,3,4), (5,6,7,8),...]

That is, each time next() is called, a tuple of the next N items is
returned.

Here's one that abuses a for loop:

from itertools import islice

def grouper(seq,n):
it = iter(seq)
for x in it:
yield (x,) + tuple(islice(it,n-1))

def test():
L = range(11)
n = 3
for x in grouper(L,n):
print x

if __name__ == '__main__':
test()

BTW what's up with the followup to gmane?

P.