built in function for computing overlap in Python

Question

is there a built in function to compute the overlap between two discrete intervals, e.g. the overlap between [10, 15] and [20, 38]? In that case the overlap is 0. If it's [10, 20], [15, 20], the overlap is 5.

Answer 1

Not the way I would ordinarily do this, but it shows the work and explains the assumptions and how it operates, and shows all the work.

def getOverlap(a:list, b:list)->int:
    # Assumptions are:
    #     both a and b are a 2-element list/vector
    #     each element of each list is an integer
    #     each list defines a range with a starting and ending point
    # There is no requirement that the starting point be greater than the ending point
    # 
    # Returns the _number_ of common elements that exist in both ranges
    #     the input lists (a and b) are unmodified
    #
    # Function  first sorts each tuple
    #           then it uses the sorted tuples to make the calculation, 
    #              which means element 0 is lower than element 1 in each tuple
    #           the elements of each tuple can be accessed by its index (0, or 1)
    #           the min() and max() functions are used to extract either the 
    #              minimum or the maximum of (here, a pair) of values
    #           the difference between the resulting min and max (plus 1) yields
    #              the number of common elements that exist in both ranges
    #              (if this number is negative, then there are no common elements)
    #           it then applies the max function to return the proper answer
    #           
    sa = sorted(a)
    sb = sorted(b)
    lowest_endpoint = min(sa[1], sb[1])
    highest_origin  = max(sa[0], sb[0])
    number_of_common_elements = lowest_endpoint - highest_origin + 1
    return max(0, number_of_common_elements)

or, more simply:

def getOverlap(a:list, b:list)->int:
    sa = sorted(a)
    sb = sorted(b)
    return max(0, min(sa[1], sb[1]) - max(sa[0], sb[0]) + 1)

Answer 2

def Overlap(self, R1, R2):
   if (A1[0]>=A2[2]) or (A1[2]<=A2[0]) or (A1[3]<=A2[1]) or (A1[1]>=A2[3]):
      return False
   else:
      return True
ob = Solution()
print(ob.Overlap([0,0,2,2],[1,1,3,3]))

Answer 3

Did some more checks to clarify behaviour & correctness:

def get_intersection_overlap(a, b):
    """
    Returns the intersection over union of two bounding boxes.
    Note, lower and upper bounds intersect exactly, it is considered not an intersection.

    ref:
        - https://stackoverflow.com/a/2953979/1601580
    """
    return max(0, min(a[1], b[1]) - max(a[0], b[0]))

def get_intersection_overlap_care_about_exact_match(a, b):
    """
    Return the amount of overlap, in bp
    between a and b.
    If >0, the number of bp of overlap
    If 0,  they are book-ended.
    If <0, the distance in bp between them

    ref:
        - https://stackoverflow.com/a/52388579/1601580
    """

    return min(a[1], b[1]) - max(a[0], b[0])

tests

def overlap_intersection_test():
    """
    want to test if two intervals intersect/overlap and return true if they do
    """
    print('----')
    print(f'{get_intersection_overlap([10, 25], [20, 38])}')
    assert get_intersection_overlap([10, 25], [20, 38]) == 5
    print(f'{get_intersection_overlap([20, 38], [10, 25])}')
    assert get_intersection_overlap([20, 38], [10, 25]) == 5

    print(f'{get_intersection_overlap([10, 15], [20, 38])}')
    assert get_intersection_overlap([10, 15], [20, 38]) == 0
    print(f'{get_intersection_overlap([10, 15], [20, 38])}')
    assert get_intersection_overlap([10, 15], [20, 38]) == 0

    print(f'{get_intersection_overlap([10, 15], [15, 38])}')
    assert get_intersection_overlap([10, 15], [15, 38]) == 0

    # -
    print('----')
    print(f'{get_intersection_overlap_care_about_exact_match([10, 25], [20, 38])}')
    assert get_intersection_overlap_care_about_exact_match([10, 25], [20, 38]) == 5
    print(f'{get_intersection_overlap_care_about_exact_match([20, 38], [10, 25])}')
    assert get_intersection_overlap_care_about_exact_match([20, 38], [10, 25]) == 5

    print(f'{get_intersection_overlap_care_about_exact_match([10, 15], [20, 38])}')
    assert get_intersection_overlap_care_about_exact_match([10, 15], [20, 38]) == -5
    print(f'{get_intersection_overlap_care_about_exact_match([10, 15], [20, 38])}')
    assert get_intersection_overlap_care_about_exact_match([10, 15], [20, 38]) == -5

    print(f'{get_intersection_overlap_care_about_exact_match([10, 15], [15, 38])}')
    assert get_intersection_overlap_care_about_exact_match([10, 15], [15, 38]) == 0

output:

----
5
5
0
0
0
----
5
5
-5
-5
0

Answer 4

I had to process inclusive bounds so the current answers did not work. Here is a solution with inclusive bounds if you only care about a True/False answer:

def overlap(first: int, last: int, another_first: int, another_last: int)->bool:
    """
    Return True if the two intervals overlap.

    >>> not any([
    ...     _overlap(1, 1, 2, 2),
    ...     _overlap(2, 2, 1, 1)
    ... ])
    True

    >>> all([
    ...     _overlap(1, 1, 1, 1),
    ...     _overlap(1, 5, 1, 1),
    ...     _overlap(1, 1, 1, 5),
    ...     _overlap(1, 3, 2, 5),
    ...     _overlap(2, 5, 1, 3),
    ...     _overlap(1, 5, 2, 3),
    ...     _overlap(2, 3, 1, 5),
    ...  ])
    True
    """
    return min(last, another_last) - max(first, another_first) >= 0

Answer 5

Just wrote this:

def overlap(interval1, interval2):
    """
    Given [0, 4] and [1, 10] returns [1, 4]
    """
    if interval2[0] <= interval1[0] <= interval2[1]:
        start = interval1[0]
    elif interval1[0] <= interval2[0] <= interval1[1]:
        start = interval2[0]
    else:
        raise Exception("Intervals are not overlapping")

    if interval2[0] <= interval1[1] <= interval2[1]:
        end = interval1[1]
    elif interval1[0] <= interval2[1] <= interval1[1]:
        end = interval2[1]
    else:
        raise Exception("Intervals are not overlapping")

    return (start, end)


def percentage_overlap(interval1, interval2):
    """
    Given [0, 4] and [1, 10] returns 0.75
    """
    try:
        overlap = _overlap(interval1, interval2)
    except Exception:
        return 0.0
    return (overlap[1] - overlap[0]) / (interval1[1] - interval1[0])

Answer 6

Here is a good function from Aaron Quinlan's chrom_sweep, modified for your interval representation. It returns the number of bp of overlap if they do overlap, otherwise it returns the distance as a negative int.

def overlaps(a, b):
    """
    Return the amount of overlap, in bp
    between a and b.
    If >0, the number of bp of overlap
    If 0,  they are book-ended.
    If <0, the distance in bp between them
    """

    return min(a[1], b[1]) - max(a[0], b[0])

Answer 7

You can use max and min:

>>> def getOverlap(a, b):
...     return max(0, min(a[1], b[1]) - max(a[0], b[0]))

>>> getOverlap([10, 25], [20, 38])
5
>>> getOverlap([10, 15], [20, 38])
0

Answer 8

Check out pyinterval http://code.google.com/p/pyinterval/

import interval
x=interval.interval[10, 15]
y=interval.interval[20, 38]
z=interval.interval[12,18]

print(x & y)
# interval()
print(x & z)
# interval([12.0, 15.0])