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Python计算两个box的IOU

Python计算两个box的IOU

作者: 几时见得清梦 | 来源:发表于2019-08-22 19:19 被阅读0次
#RT:RightTop
#LB:LeftBottom
def IOU(rectangle A, rectangleB):
    W = min(A.RT.x, B.RT.x) - max(A.LB.x, B.LB.x)
    H = min(A.RT.y, B.RT.y) - max(A.LB.y, B.LB.y)
    if W <= 0 or H <= 0:
        return 0;
    SA = (A.RT.x - A.LB.x) * (A.RT.y - A.LB.y)
    SB = (B.RT.x - B.LB.x) * (B.RT.y - B.LB.y)
    cross = W * H
    return cross/(SA + SB - cross)

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