soft-nms

Posted 2020-11-21 ymjyqsx

https://blog.csdn.net/shuzfan/article/details/71036040

github：https://github.com/bharatsingh430/soft-nms

解决的问题：就是两个框iou有一定重叠且两个框的得分都很高（同时两个框确实包含了我们想要的检测结果），这样有一个框会被nms过滤掉

解决的方法：之前的nms是直接把低分框过滤掉（或者按照论文说的把低分框的score置为0），现在是把低分框的得分降低，具体有两种降低方式

 

在lib/nms/cpu_nms.pyx

值得注意的是：iou的阈值是0.3，不是0.5，论文里面说好像是做实验对比的几个检测器也是用的0.3的阈值

def cpu_soft_nms(np.ndarray[float, ndim=2] boxes, float sigma=0.5, float Nt=0.3, float threshold=0.001, unsigned int method=0):
    cdef unsigned int N = boxes.shape[0]
    cdef float iw, ih, box_area
    cdef float ua
    cdef int pos = 0
    cdef float maxscore = 0
    cdef int maxpos = 0
    cdef float x1,x2,y1,y2,tx1,tx2,ty1,ty2,ts,area,weight,ov

    for i in range(N):　　　　　　　　　　　　　　每次找最大的得分和相应的box
        maxscore = boxes[i, 4]
        maxpos = i

        tx1 = boxes[i,0]
        ty1 = boxes[i,1]
        tx2 = boxes[i,2]
        ty2 = boxes[i,3]
        ts = boxes[i,4]

        pos = i + 1
    # get max box
        while pos < N:
            if maxscore < boxes[pos, 4]:
                maxscore = boxes[pos, 4]
                maxpos = pos
            pos = pos + 1

    # add max box as a detection 
        boxes[i,0] = boxes[maxpos,0]
        boxes[i,1] = boxes[maxpos,1]
        boxes[i,2] = boxes[maxpos,2]
        boxes[i,3] = boxes[maxpos,3]
        boxes[i,4] = boxes[maxpos,4]

    # swap ith box with position of max box　　　　　　把得分最大的放到当前第一个位置
        boxes[maxpos,0] = tx1
        boxes[maxpos,1] = ty1
        boxes[maxpos,2] = tx2
        boxes[maxpos,3] = ty2
        boxes[maxpos,4] = ts

        tx1 = boxes[i,0]
        ty1 = boxes[i,1]
        tx2 = boxes[i,2]
        ty2 = boxes[i,3]
        ts = boxes[i,4]

        pos = i + 1
    # NMS iterations, note that N changes if detection boxes fall below threshold
        while pos < N:　　　　　　　　　　　　　　　　　　　　　　当前第一个，也就是得分最高的一个，和后面所有的box进行nms操作
            x1 = boxes[pos, 0]
            y1 = boxes[pos, 1]
            x2 = boxes[pos, 2]
            y2 = boxes[pos, 3]
            s = boxes[pos, 4]

            area = (x2 - x1 + 1) * (y2 - y1 + 1)
            iw = (min(tx2, x2) - max(tx1, x1) + 1)　　　　　　width的重叠部分长度
            if iw > 0:
                ih = (min(ty2, y2) - max(ty1, y1) + 1)　　　 height的重叠部分长度
                if ih > 0:
                    ua = float((tx2 - tx1 + 1) * (ty2 - ty1 + 1) + area - iw * ih)
                    ov = iw * ih / ua #iou between max box and detection box

                    if method == 1: # linear
                        if ov > Nt: 
                            weight = 1 - ov
                        else:
                            weight = 1
                    elif method == 2: # gaussian
                        weight = np.exp(-(ov * ov)/sigma)
                    else: # original NMS
                        if ov > Nt: 
                            weight = 0
                        else:
                            weight = 1

                    boxes[pos, 4] = weight*boxes[pos, 4]
            
            # if box score falls below threshold, discard the box by swapping with last box
            # update N
                    if boxes[pos, 4] < threshold:
                        boxes[pos,0] = boxes[N-1, 0]
                        boxes[pos,1] = boxes[N-1, 1]
                        boxes[pos,2] = boxes[N-1, 2]
                        boxes[pos,3] = boxes[N-1, 3]
                        boxes[pos,4] = boxes[N-1, 4]
                        N = N - 1
                        pos = pos - 1

            pos = pos + 1

    keep = [i for i in range(N)]
    return keep

核心部分，实际上就改了这部分：

                    if method == 1: # linear
                        if ov > Nt: 
                            weight = 1 - ov
                        else:
                            weight = 1
                    elif method == 2: # gaussian
                        weight = np.exp(-(ov * ov)/sigma)
                    else: # original NMS
                        if ov > Nt: 
                            weight = 0
                        else:
                            weight = 1

线性衰减法：（1-overlap）×之前的得分  =  现在的得分

高斯衰减发：-overlap的平方/0.5，然后开e次方