interval overlap size

Question

what is the most concise way in Python to compute interval overlap size?

overlap([a, b], [c, d]) should return 0 if intervals are identical, N when they overlap but not identical (where N is overlap) and None if they are not overlapping.

thanks.

edit: overlap is misleading i mean the size by which the intervals are non-overlapping. so 0 is they are identical.

http://stackoverflow.com/questions/2953967/built-in-function-for-computing-overlap-in-python ? — sjr, Jun 14 '12 at 03:19
Why would you want to return 0 if the intervals are identical? That seems inconsistent with returning N when the intervals overlap but aren't identical. — Li-aung Yip, Jun 14 '12 at 03:21
Please give an example with some overlap. It will clarify the question. — damned, Jun 14 '12 at 03:24
Please clear up if N is size of overlap or size of non-overlap. — Reinstate Monica, Jun 14 '12 at 06:50
What if the intervals touch, i.e. what should `overlap([1, 2], [2, 4])` return? — Reinstate Monica, Jun 14 '12 at 14:35

Junuxx · Answer 1 · 2012-06-14T10:16:40.983

3

It doesn't get much more concise than the accepted answer in the question linked to by sjr, but:

def overlap(a,b,c,d):
    r = 0 if a==c and b==d else min(b,d)-max(a,c)
    if r>=0: return r

would also return 0 for identical intervals and None for non-overlapping intervals as required.

edited Jun 14 '12 at 10:16

answered Jun 14 '12 at 03:29

Junuxx

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0 should be returned for identical intervals, not `None`. Also, your function, as written, will return `None` for any input. – Reinstate Monica Jun 14 '12 at 06:45

interval overlap size

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