def iou(a, b):
"""
:param a:4*M*1 left,top,right,bottom
:param b:4*1*N left,top,right,bottom
:return:
"""
aleft, atop, aright, abottom = [a[i] for i in range(4)]
bleft, btop, bright, bbottom = [b[i] for i in range(4)]
cross_left = np.maximum(aleft, bleft)
cross_top = np.maximum(atop, btop)
cross_right = np.minimum(aright, bright)
cross_bottom = np.minimum(abottom, bbottom)
# 如果没有交集就需要裁切(M*N)
cross_area = (cross_right - cross_left + 1).clip(0) * (cross_bottom - cross_top + 1).clip(0)
union_area = (aright - aleft + 1) * (abottom - atop + 1) + (bright - bleft + 1) * (bbottom - btop + 1) - cross_area
return cross_area / union_area
if __name__ == '__main__':
a = np.arange(0, 12).reshape(4, -1, 1)
b = np.arange(12, 24).reshape(4, 1, -1)
iou(a,b)
标签：__,right,area,批次,iou,bottom,cross,np From： https://www.cnblogs.com/xiaoruirui/p/16965299.html