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  • 【NMS与IOU代码】

    # -*- coding: utf-8 -*-
    import numpy as np
    def IOU1(A,B):
        #左上右下坐标(x1,y1,x2,y2)
        w=max(0,min(A[2],B[2])-max(A[0],B[0]))
        h=max(0,min(A[3],B[3])-max(A[1],B[1]))
        areaA=(A[2]-A[0]+1)*(A[3]-A[1]+1)
        areaB=(B[2]-B[0]+1)*(B[3]-B[1]+1)
        inter=w*h
        union=areaA+areaB-inter
        return inter/union
    def nms(dets, thresh):
        """Pure Python NMS baseline."""
        #x1、y1、x2、y2、以及score赋值
        x1 = dets[:, 0]
        y1 = dets[:, 1]
        x2 = dets[:, 2]
        y2 = dets[:, 3]
        scores = dets[:, 4]
        areas = (x2 - x1 + 1) * (y2 - y1 + 1)
        order = scores.argsort()[::-1]
    
        keep = []
        while order.size > 0:#还有数据
            i = order[0]
            keep.append(i)
            #计算当前概率最大矩形框与其他矩形框的相交框的坐标
            xx1 = np.maximum(x1[i], x1[order[1:]])
            yy1 = np.maximum(y1[i], y1[order[1:]])
            xx2 = np.minimum(x2[i], x2[order[1:]])
            yy2 = np.minimum(y2[i], y2[order[1:]])
    
            #计算相交框的面积
            w = np.maximum(0.0, xx2 - xx1 + 1)
            h = np.maximum(0.0, yy2 - yy1 + 1)
            inter = w * h
            #计算重叠度IOU:重叠面积/(面积1+面积2-重叠面积)
            IOU = inter / (areas[i] + areas[order[1:]] - inter)
            #找到重叠度不高于阈值的矩形框索引
            left_index = np.where(IOU <= thresh)[0]
            #将order序列更新,由于前面得到的矩形框索引要比矩形框在原order序列中的索引小1,所以要把这个1加回来
            order = order[left_index + 1]
        print(keep)
    if __name__ == '__main__':
         dets=[[0,0,100,101,0.9],[5,6,90,110,0.7],[17,19,80,120,0.8],[10,8,115,105,0.5]]
         dets=np.array(dets)
         nms(dets,0.5)
         print IOU1(dets[0],dets[2])
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  • 原文地址:https://www.cnblogs.com/EstherLjy/p/9392170.html
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