Calculate overlapped area between two rectangles

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

Since the post is very related to computer vision and object detection, I thought of putting some code together that I use for finding the intersection of bounding boxes and also finding their intersection over union (IoU). This code was originally developed by Adrian Rosebrock in this blog post:

This is the module (where I named it Bbox):

class Bbox:
    def __init__(self, x1, y1, x2, y2):
        self.x1 = max(x1, x2)
        self.x2 = min(x1, x2)
        self.y1 = max(y1, y2)
        self.y2 = max(y1, y2)
        self.box = [self.x1, self.y1, self.x2, self.y2]
        self.width = abs(self.x1 - self.x2)
        self.height = abs(self.y1 - self.y2)

    @property
    def area(self):
        """
        Calculates the surface area. useful for IOU!
        """
        return (self.x2 - self.x1 + 1) * (self.y2 - self.y1 + 1)

    def intersect(self, bbox):
        x1 = max(self.x1, bbox.x1)
        y1 = max(self.y1, bbox.y1)
        x2 = min(self.x2, bbox.x2)
        y2 = min(self.y2, bbox.y2)
        intersection = max(0, x2 - x1 + 1) * max(0, y2 - y1 + 1)
        return intersection

    def iou(self, bbox):
        intersection = self.intersection(bbox)

        iou = intersection / float(self.area + bbox.area - intersection)
        # return the intersection over union value
        return iou

And to use it:

a = Bbox([516, 289, 529, 303])
b = Bbox([487, 219, 533, 342])

result = a.intersect(b)

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.