【代码】NMS_nms python-优快云博客

本文链接：https://blog.youkuaiyun.com/FSALICEALEX/article/details/90261346

本文详细解析了FastR-CNN中使用的非极大值抑制(NMS)算法，通过Python实现，介绍了如何计算边界框的交并比(IOU)，并展示了NMS算法的具体流程。

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# --------------------------------------------------------
# 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