# --------------------------------------------------------
# Fast R-CNN
# Copyright (c) 2015 Microsoft
# Licensed under The MIT License [see LICENSE for details]
# Written by Ross Girshick
# --------------------------------------------------------
import numpy as np
def py_cpu_nms(dets, thresh):
"""Pure Python NMS baseline."""
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:
i = order[0]
keep.append(i)
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:]])
w = np.maximum(0.0, xx2 - xx1 + 1)
h = np.maximum(0.0, yy2 - yy1 + 1)
inter = w * h
ovr = inter / (areas[i] + areas[order[1:]] - inter)
inds = np.where(ovr <= thresh)[0]
order = order[inds + 1]
return keep
以上为代码。下面说说它是怎么来的。
假设det1为[a1, b1, a2, b2], det2为[x1, y1, x2, y2],我们要求他们的交和并。
det1的集合表示为{(x, y) | a1 < x < a2, b1 < y < b2},det2 = {(x, y) | x1 < x < x2, y1 < y < y2}
求交
求两个集合的交Int = {(x, y) | a1 < x < a2 且 b1 < y < b2 且 x1 < x < x2 且 y1 < y < y2}
= {(x, y) | a1 < x < a2 且 x1 < x < x2 且 b1 < y < b2 且 y1 < y < y2}
= {(x, y) | max(a1, x1) < x < min(a2, x2), max(b1, y1) < y < min(b2, y2)}
其中要注意的是可能会出现max(a1, x1) > min(a2, x2)这种情况出现,这时Int为空集。
求并
两个集合的并不一定是矩形,但面积可以通过交来算出来
NMS
keep=[]
score降序排列
while(score不为空)
A = score最大的det
将其余det与A计算IOU
删除IOU过小的det
score中删除A
keep中添加A
返回keep