Split a python list into other "sublists" i.e smaller lists [duplicate]

Question

I'd say

chunks = [data[x:x+100] for x in range(0, len(data), 100)]

If you are using python 2.x instead of 3.x, you can be more memory-efficient by using xrange(), changing the above code to:

chunks = [data[x:x+100] for x in xrange(0, len(data), 100)]

Actually I think using plain slices is the best solution in this case:

for i in range(0, len(data), 100):
    chunk = data[i:i + 100]
    ...

If you want to avoid copying the slices, you could use itertools.islice(), but it doesn't seem to be necessary here.

The itertools() documentation also contains the famous "grouper" pattern:

def grouper(n, iterable, fillvalue=None):
    "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

You would need to modify it to treat the last chunk correctly, so I think the straight-forward solution using plain slices is preferable.

chunks = [data[100*i:100*(i+1)] for i in range(len(data)/100 + 1)]

This is equivalent to the accepted answer. For example, shortening to batches of 10 for readability:

data = range(35)
print [data[x:x+10] for x in xrange(0, len(data), 10)]
print [data[10*i:10*(i+1)] for i in range(len(data)/10 + 1)]

Outputs:

[[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], [30, 31, 32, 33, 34]]
[[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], [30, 31, 32, 33, 34]]