python实现的Iou与Giou

最近看了网上很多博主写的iou实现方法,但Giou的代码似乎比较少,于是便自己写了一个,新手上路,如有错误请指正,话不多说,上代码:

def Iou(rec1,rec2):
    x1,x2,y1,y2 = rec1 #分别是第一个矩形左右上下的坐标
    x3,x4,y3,y4 = rec2 #分别是第二个矩形左右上下的坐标
    area_1 = (x2-x1)*(y1-y2)
    area_2 = (x4-x3)*(y3-y4)
    sum_area = area_1 + area_2
    w1 = x2 - x1#第一个矩形的宽
    w2 = x4 - x3#第二个矩形的宽
    h1 = y1 - y2
    h2 = y3 - y4
    W = min(x1,x2,x3,x4)+w1+w2-max(x1,x2,x3,x4)#交叉部分的宽
    H = min(y1,y2,y3,y4)+h1+h2-max(y1,y2,y3,y4)#交叉部分的高
    Area = W*H#交叉的面积
    Iou = Area/(sum_area-Area)
    return Iou

def Giou(rec1,rec2):
    x1,x2,y1,y2 = rec1 #分别是第一个矩形左右上下的坐标
    x3,x4,y3,y4 = rec2
    iou = Iou(rec1,rec2)
    area_C = (max(x1,x2,x3,x4)-min(x1,x2,x3,x4))*(max(y1,y2,y3,y4)-min(y1,y2,y3,y4))
    area_1 = (x2-x1)*(y1-y2)
    area_2 = (x4-x3)*(y3-y4)
    sum_area = area_1 + area_2
    w1 = x2 - x1#第一个矩形的宽
    w2 = x4 - x3#第二个矩形的宽
    h1 = y1 - y2
    h2 = y3 - y4
    W = min(x1,x2,x3,x4)+w1+w2-max(x1,x2,x3,x4)#交叉部分的宽
    H = min(y1,y2,y3,y4)+h1+h2-max(y1,y2,y3,y4)#交叉部分的高
    Area = W*H#交叉的面积
    add_area = sum_area - Area #两矩形并集的面积
    end_area = (area_C - add_area)/area_C #(c/(AUB))/c的面积
    giou = iou - end_area
    return giou


rec1 = (27,47,130,90)
rec2 = (30,68,150,110)
iou = Iou(rec1,rec2)
giou = Giou(rec1,rec2)
print("Iou = {},Giou = {}".format(iou,giou))

关于giou的具体内容可以参考论文及网上一些大神的博客。

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