Find consecutive combinations [duplicate]

Question

Possible Duplicate:
Rolling or sliding window iterator in Python

I'm new to programming and am learning Python. I'm looking for an efficient/pythonic way to solve a problem.

I'd like a function that returns a list of iterables containing the combinations of a parent iterable as long as the elements in the combination appear the same same consecutive order as the original parent iterable.

I'm not sure if "consecutive" if the right word to describe this concept as 'consecutive' typically means, 'the same element repeated.' e.g. [1,1,1], 'aaa', etc...

I mean that given the list [1,2,3,4,5]:

[1,2,3] is consecutive but [1,2,4] is not. (Is there a word for this?)

Here's a function consecutive_combinations() I created and the expected behavior:

def consecutive_combinations(iterable, consec):
    begin = 0
    chunks = len(iterable) + 1 - consec
    return [iterable[x + begin: x + consec] for x in xrange(chunks)]

def test():
    t = (1,2,3,4,5)
    s = "The quick brown fox jumps over the lazy dog."
    CC = consecutive_combinations
    assert CC(t, 2) == [(1, 2), (2, 3), (3, 4), (4, 5)]
    assert CC(t, 3) == [(1, 2, 3), (2, 3, 4), (3, 4, 5)]
    assert CC(t, 4) == [(1, 2, 3, 4), (2, 3, 4, 5)]
    assert CC(t, 5) == [(1, 2, 3, 4, 5)]
    assert CC(s, 3) == ['The', 'he ', 'e q', ' qu', 'qui', 'uic', 'ick', 'ck ', 'k b', ' br', 'bro', 'row', 'own', 'wn ', 'n f', ' fo', 'fox', 'ox ', 'x j', '  ju', 'jum', 'ump', 'mps', 'ps ', 's o', ' ov', 'ove', 'ver', 'er ', 'r t', '    th', 'the', 'he ', 'e l', ' la', 'laz', 'azy', 'zy ', 'y d', ' do', 'dog', 'og. ']
    assert CC('', 3) == []
    print "All tests passed!"

test()

Is this an efficient solution? Is there something in itertools or some other pre-built module that would do this sort of thing?

Answer 1

I like the pragmatic zip approach:

n = 3
s = "The quick brown fox jumps over the lazy dog."
zip(*(s[i:] for i in xrange(n)))

It's not super-efficient and it only works for sequences, but often enough it does the job.

The corresponding itertools solution is a pretty straightforward transformation of the above:

from itertools import izip, islice, tee

def slices(iterable, n):
    return izip(*(islice(it, i, None) for i, it in enumerate(tee(iterable, n))))

Soo many is...

Nevertheless, this one should work for any iterable (but may be slower for plain sequences like lists or strings).

Answer 2

Your solution is fine. You could however shorten it a little bit. For example:

def subsequences(iterable, length):
    return [iterable[i: i + length] for i in xrange(len(iterable) - length + 1)]