Intersection over Union

Compute the IoU of two axis-aligned bounding boxes.

Each box is [x1, y1, x2, y2] with (x1, y1) the top-left corner and (x2, y2) the bottom-right.

IoU = area of intersection / area of union

The intersection rectangle spans

x1 = max(a.x1, b.x1)   y1 = max(a.y1, b.y1)
x2 = min(a.x2, b.x2)   y2 = min(a.y2, b.y2)

and its width and height are max(0, x2 - x1) and max(0, y2 - y1). The clamp at zero is the whole problem: without it, two boxes that miss each other produce a negative width times a negative height, which is a positive "overlap" and a detector that scores best when it is most wrong.

The union is area_a + area_b - intersection. If it is zero, return 0.0.

IoU is the threshold that decides whether a detection counts as a hit, usually at 0.5, and it is what non-maximum suppression uses to drop duplicate boxes.

Function signature

def iou(box_a, box_b):

Example

Input

iou([0, 0, 2, 2], [1, 1, 3, 3])

Output

0.14285714285714285

Reasoning

The boxes overlap in the square from (1, 1) to (2, 2), so the intersection is 1. Each box has area 4, so the union is 4 + 4 - 1 = 7, and IoU is 1/7.

Constraints and notes

• Clamp the intersection width and height at 0.
• Return 0.0 when the union is 0.