How to chunk up an interval in Python?

Question

Is there a built-in/existing library function that is like xrange but splits an interval into even spaced non-overlapping chunks?

For instance, if we call this function xchunks, then I would like:

>>> xchunks(start=0, stop=18, step=5)
[(0, 4), (5, 9), (10, 14), (15, 17)]

Ideally, this should also work for negative step.

>>> xchunks(start=20, stop=2, step=5)
[(20, 16), (15, 11), (10, 6), (5, 3)]

Answer 1

This (almost) does what you want:

[(s, s + step - 1) for s in range(start, stop, step)]

Result:

[(0, 4), (5, 9), (10, 14), (15, 19)]

You can improve on this an make the last pair (15, 17).

Answer 2

This gives an iterator as with xrange and also works with a negative step. If you're using Python 3.x then replace xrange with range.

def xchunks(start, stop, step):
    return ((i, min(i + step - 1, stop - 1)) if step > 0 else
            (i, max(i + step + 1, stop + 1))
            for i in xrange(start, stop, step))


def test():
    result = list(xchunks(start=0, stop=18, step=5))
    print list(result)
    assert result == [(0, 4), (5, 9), (10, 14), (15, 17)]

    result = list(xchunks(start=20, stop=2, step=-5))
    print list(result)
    assert result == [(20, 16), (15, 11), (10, 6), (5, 3)]

Answer 3

the full solution to this would look like:

[(s, (s+step-1 if s+step-1<stop-1 else stop-1)) for s in xrange(start,stop,step)]

use range or xrange, whichever you like.