深度学习八—— 手撕 IoU

8. 手撕IoU

矩形框的IoU计算

矩阵$\text{T}$的左下角坐标为 $(X_0,Y_0)$，右上角坐标为$(X_1,Y_1)$

矩阵$\text{G}$的左下角坐标为 $(A_0,B_0)$，右上角坐标为$(A_1,B_1)$

则IoU的计算过程如下：

$$IoU=\frac{area(RO{I}_T)\cap area(RO{I}_G)}{S_{XY}+S_{AB}-area(RO{I}_T)\cap area(RO{I}_G)}$$

手撕 IoU

# A:
# 左下角坐标(left_x,left_y)
# 右上角坐标(right_x,right_y)
# B:
# 左下角坐标(left_x,left_y)
# 右上角坐标(right_x,right_y)
 
def IOU(rectangle A, rectangleB):
    W = min(A.right_x, B.right_x) - max(A.left_x, B.left_x)
    H = min(A.right_y, B.right_y) - max(A.left_y, B.left_y)
    if W <= 0 or H <= 0:
        return 0;
    SA = (A.right_x - A.left_x) * (A.right_y - A.left_y)
    SB = (B.right_x - B.left_x) * (B.right_y - B.left_y)
    cross = W * H
    return cross/(SA + SB - cross)