Most pythonic way to group an iterator in blocks of size n

Question

Possible Duplicate:
How do you split a list into evenly sized chunks in Python?

If you have an iterator iter with values i0, i1, i2, ..., what's the most Pythonic way to get an iterator whose values are

[i0, i1, i2], [i3, i4, i5], ...

The above would be the output to a hypothetical groupby(iter, 3).

More specifically, [this answer](http://stackoverflow.com/a/312644/989121) which applies to arbitrary iterables. — georg, Aug 22 '12 at 14:57

score 2 · Answer 1 · answered Aug 22 '12 at 14:42

2

The itertools module has a recipe for this called grouper. Here is the excerpt copied straight from itertools documentation.

def grouper(n, iterable, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

answered Aug 22 '12 at 14:42

Praveen Gollakota

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Coming to think of it, it's exactly similar to the solution provided by @user792036 – Praveen Gollakota Aug 22 '12 at 14:44
Actually this covers a fill value and looks nicer when used since it's a function. – jamylak Aug 22 '12 at 15:46

score 0 · Answer 2 · answered Aug 22 '12 at 14:37

0

The best solution that I can find (requires itertools):

groupby = lambda s,n : izip(*[iter(s)]*n)

but it feels slightly hacky and is hard to read

answered Aug 22 '12 at 14:37

user792036

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That does not solve the problem in a general way for inputs that are not a multiple of the block size. izip basically ignores the last, not full, block of items. – yacc143 Nov 06 '14 at 10:12

Most pythonic way to group an iterator in blocks of size n

2 Answers2