https://blog.csdn.net/shuzfan/article/details/71036040
github：https://github.com/bharatsingh430/soft-nms，代码在/lib/nms/下
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
nms 2种衰减法和原始的nms方式：
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
soft-nms如何过滤掉框：如果经过衰减后的框的得分小于阈值，把最后一个位置的框和这个框交换位置，然后N-1，相当于最后不会遍历到这个框(外层循环和内层循环都不会遍历)，也就过滤掉了。阈值的设定为threshold=0.001，实际上我觉得这个值设的有些小，当然他的衰减本身有点多
# 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