两个不规则的box求取IOU-python

第一步：构造该框

以四条线段表示该框，构造一个二进制的mask
因为框里的像素只需满足四个公式，在L4的下面，在L2的上面，在L1的左边，在L3的右边

第二步：相与和相或

两框的mask相与求交集，相或求并集，二者相除得到iou

import os
import numpy as np
def get_k(x1,y1,x2,y2,x3,y3,x4,y4):
    k1=(y2-y1)/(x2-x1)
    k2=(y3-y2)/(x3-x2)
    k3=(y4-y3)/(x4-x3)
    k4=(y4-y1)/(x4-x1)
    bias1=y1-k1*x1
    bias2 = y2 - k2 * x2
    bias3 = y3 - k3 * x3
    bias4 = y4 - k4 * x4
    return k1,k2,k3,k4,bias1,bias2,bias3,bias4
def get_area(list1,width,height):
    x1, y1, x2, y2, x3, y3, x4, y4=list1
    k1,k2,k3,k4,bias1,bias2,bias3,bias4=get_k(x1,y1,x2,y2,x3,y3,x4,y4)

    x_index=np.arange(0,width)
    x_index=x_index.repeat(height,axis=-1).reshape(width,height).transpose()
    y_index=np.arange(0,height)
    y_index = y_index.repeat(width, axis=-1).reshape(height, width)
    L1_statisfied=((y_index -bias1) /k1 > x_index)
    L2_statisfied = (k2 * x_index + bias2 - y_index < 0)
    L3_statisfied = ((y_index -bias3) /k3 < x_index)
    L4_statisfied = (k4 * x_index + bias4 - y_index > 0)
    image=L1_statisfied&L2_statisfied&L3_statisfied&L4_statisfied
    return image
def get_iou(list1,list2):
    width=max(list1[0],list1[2],list1[4],list1[6],list2[0],list2[2],list2[4],list2[6])
    height=max(list1[1],list1[3],list1[5],list1[7],list2[1],list2[3],list2[5],list2[7])
    image1=get_area(list1,width,height)
    image2=get_area(list2,width,height)
    jiaoji=image1&image2
    jiaoji=np.array(jiaoji).astype(np.int)
    bingji=image1|image2
    bingji = np.array(bingji).astype(np.int)
    iou=jiaoji.sum()/bingji.sum()
    return iou
list1=[4,5,6,2,1,1,2,3]
list2=[4,5,6,2,1,1,2,3]
iou=get_iou(list1,list2)
print(iou)