Code golfing: splitting a list

[Ofnuts]

So I have a list L. To do some parallel processing, I want to split it into N sublists of equivalent length. The contents of the sublists are indifferent. What is your shortest and/or most pythonic code? Can you avoid using len(L) explicitly? 

[ichabod801]

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M = [L[start::N] for start in range(N)]

Assumes N divides evenly into len(L). If that's a concern you could put a second index into the slice double zip it to account for that.

[Mekire]

So you want to chunk?
Lots of good answers here:
http://stackoverflow.com/questions/31244...zed-chunks

The standard idiom people generally use is:

chunks = (L[i:i + N] for i in range(0, len(L), N))

This assumes the whole sequence fits in memory which it sounds like you are trying to avoid.

Alternatives use iter and zip and look like this:

chunks_zip = zip(*[iter(L)]*N)

The second version above ignores groups that are not full.  But can handle generators that wouldn't otherwise fit in memory:
Python3

L = range(10**25)
N = 3

chunks_zip = zip(*[iter(L)]*N)

for i in range(10):
    print(next(chunks_zip))

Output:
(0, 1, 2)
(3, 4, 5)
(6, 7, 8)
(9, 10, 11)
(12, 13, 14)
(15, 16, 17)
(18, 19, 20)
(21, 22, 23)
(24, 25, 26)
(27, 28, 29)

Use zip_longest with a fill value if you don't want to lose incomplete groups.

[Ofnuts]

Hide/Show

M = [L[start::N] for start in range(N)]

Assumes N divides evenly into len(L). If that's a concern you could put a second index into the slice double zip it to account for that.

This is more or less what my code does, but with slightly clumsier Python. And as far as I can tell this works even if N is doesn't divide len(N).

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So you want to chunk?
Lots of good answers here:
http://stackoverflow.com/questions/31244...zed-chunks

The standard idiom people generally use is:

chunks = (L[i:i + N] for i in range(0, len(L), N))

This assumes the whole sequence fits in memory which it sounds like you are trying to avoid.

No, in the real-life problem I am handling rather small lists, it just that each item ends up used in a webservice call so to speed things up I create N threads and give each a part of the original list.

Alternatives use iter and zip and look like this:

chunks_zip = zip(*[iter(L)]*N)

The second version above ignores groups that are not full.  But can handle generators that wouldn't otherwise fit in memory:
Python3

L = range(10**25)
N = 3

chunks_zip = zip(*[iter(L)]*N)

for i in range(10):
    print(next(chunks_zip))

Output:
(0, 1, 2)
(3, 4, 5)
(6, 7, 8)
(9, 10, 11)
(12, 13, 14)
(15, 16, 17)
(18, 19, 20)
(21, 22, 23)
(24, 25, 26)
(27, 28, 29)

Use zip_longest with a fill value if you don't want to lose incomplete groups.

Clever, but if you are unlucky, one of the chunks produced by the idioms above can have a length of 1 so the maximum size difference is N-1. With the 1 every N sampling you get more uniform sizes (but possibly using more CPU/memory, which isn't really a concern for me but could be for someone else).

[wavic]

There is an example in the Python documentation: https://docs.python.org/3.5/library/iter...ls-recipes

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

[Ofnuts]

There is an example in the Python documentation: https://docs.python.org/3.5/library/iter...ls-recipes

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

Yes, this is Mekire's suggestion...

[ichabod801]

This is more or less what my code does, but with slightly clumsier Python. And as far as I can tell this works even if N is doesn't divide len(N).

You said you wanted lists of equivalent length. I took that to mean equal length. If N does not divide evenly into len(L), then some of the lists will be one item longer. The fixes I proposed would make them all the same length, but they would drop some items.

[wavic]

How about this

It looks simple enough for everyone.