Calculate overlapped area between two rectangles

Question

I want to calculate the overlapped area \"THE GRAY REGION\" between red and bl

Answer 1

As this question has a shapely tag, here is a solution using it. I will use the same rectangles as in the tom10 answer:

from shapely.geometry import Polygon

polygon = Polygon([(3, 3), (5, 3), (5, 5), (3, 5)])
other_polygon = Polygon([(1, 1), (4, 1), (4, 3.5), (1, 3.5)])
intersection = polygon.intersection(other_polygon)
print(intersection.area)
# 0.5

This is much more concise than the version in the accepted answer. You don't have to construct your own Rectangle class as Shapely already provides the ready ones. It's less error-prone (go figure out the logic in that area function). And the code itself is self-explanatory.

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References:
Docs for object.intersection(other) method

Answer 2

This type of intersection is easily done by the "min of the maxes" and "max of the mins" idea. To write it out one needs a specific notion for the rectangle, and, just to make things clear I'll use a namedtuple:

from collections import namedtuple
Rectangle = namedtuple('Rectangle', 'xmin ymin xmax ymax')

ra = Rectangle(3., 3., 5., 5.)
rb = Rectangle(1., 1., 4., 3.5)
# intersection here is (3, 3, 4, 3.5), or an area of 1*.5=.5

def area(a, b):  # returns None if rectangles don't intersect
    dx = min(a.xmax, b.xmax) - max(a.xmin, b.xmin)
    dy = min(a.ymax, b.ymax) - max(a.ymin, b.ymin)
    if (dx>=0) and (dy>=0):
        return dx*dy

print area(ra, rb)
#  0.5 

If you don't like the namedtuple notation, you could just use:

dx = max(a[0], b[0]) - min(a[2], b[2])

etc, or whatever notation you prefer.