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Splitting a list into even size chunks in python?

0 votes
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Hi,

I have a list of arbitrary length, and I need to split it up into equal size chunks. There are some obvious ways to do this, like keeping a counter and two lists, and when the second list fills up, add it to the first list and empty the second list for the next round of data, but this is potentially extremely expensive.

I was wondering if anyone had a good solution to this for lists of any length

This should work:

l = range(1, 1000)
print chunks(l, 10) -> [ [ 1..10 ], [ 11..20 ], .., [ 991..999 ] ]

I was looking for something useful in itertools but I couldn't find anything obviously useful.

Appretiate your help.

posted Mar 27, 2013 by Salil Agrawal

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Also - what is the chunk size? Do you know it in advance? Is it fixed or calculated from the data?

1 Answer

0 votes

Look again, for the grouper() recipe. For lists you can also use slicing:

>>> items
['a', 'b', 'c', 'd', 'e', 'f', 'g']
>>> n = 3
>>> [items[start:start+n] for start in range(0, len(items), n)]
[['a', 'b', 'c'], ['d', 'e', 'f'], ['g']]
answer Mar 27, 2013 by anonymous
Grouper() is good tty.cooked() with just a little time.time() and crypt.salt()
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