Python计算两个box的IOU
2019-08-22 本文已影响0人
几时见得清梦
#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)
