Chunks of equal size

To partition a sequence into chunks of equal size, use chop:

>>> from split import chop
>>> list(chop(3, range(10)))
[[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]]

If truncate=True keyword argument is given, then sequence length is truncated to a multiple of chunk size, and all chunks have the same size:

>>> list(chop(3, range(10), truncate=True))
[[0, 1, 2], [3, 4, 5], [6, 7, 8]]

Subsequences by a predicate

To split a sequence into two by a given predicate, use partition:

>>> from split import partition
>>> def odd(x): return x%2
>>> map(list, partition(odd, range(5)))
[[1, 3], [0, 2, 4]]

For more general partitioning, use groupby:

>>> [(k, list(i)) for k,i in groupby(lambda x: x%3, range(7))]
[(0, [0, 3, 6]), (1, [1, 4]), (2, [2, 5])]

This function is different from itertools.groupby: it returns only one subsequence iterator per predicate value. Its return value can be converted into dictionary.

When working with very long sequences, consider using predicate_values keyword argument to avoid scanning the entire sequence. For example:

>>> longseq = xrange(int(1e9))
>>> pred = lambda x: x%3
>>> dict(groupby(pred, longseq, predicate_values=(0,1,2)))
{0: <generator object subsequence at 0x301b7d0>,
 1: <generator object subsequence at 0x301b780>,
 2: <generator object subsequence at 0x301b730>}

Breaking on separators

To break a sequence into chunks on some separators, use split. For example, breaking on zero elements:

>>> list(split(0, [1,2,3,0,4,5,0,0,6]))
[[1, 2, 3], [4, 5], [], [6]]

You can use a function as a predicate too:

>>> list(split(lambda x: x==5, range(10)))
[[0, 1, 2, 3, 4], [6, 7, 8, 9]]