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非极大值抑制(NMS)

时间:2018-01-14 15:55:46      阅读:182      评论:0      收藏:0      [点我收藏+]

标签:value   minimum   pre   numpy   nat   def   搜索   就是   oat   

非极大值抑制顾名思义就是抑制不是极大值的元素,搜索局部的极大值。这个局部代表的是一个邻域,邻域有两个参数可变,一个是邻域的维数,二是邻域的大小。这里不讨论通用的NMS算法,而是用于在目标检测中提取分数最高的窗口的。例如在行人检测中,滑动窗口经提取特征,经分类器分类识别后,每个窗口都会得到一个分数。但是滑动窗口会导致很多窗口与其他窗口存在包含或者大部分交叉的情况。这时就需要用到NMS来选取那些邻域里分数最高,并且抑制那些分数低的窗口。

# import the necessary packages  
import numpy as np  
   
# Malisiewicz et al.  
def non_max_suppression_fast(boxes, overlapThresh):  
    # if there are no boxes, return an empty list  
    if len(boxes) == 0:  
        return []  
   
    # if the bounding boxes integers, convert them to floats --  
    # this is important since we‘ll be doing a bunch of divisions  
    if boxes.dtype.kind == "i":  
        boxes = boxes.astype("float")  
   
    # initialize the list of picked indexes   
    pick = []  
   
    # grab the coordinates of the bounding boxes  
    x1 = boxes[:,0]  
    y1 = boxes[:,1]  
    x2 = boxes[:,2]  
    y2 = boxes[:,3]  
   
    # compute the area of the bounding boxes and sort the bounding  
    # boxes by the bottom-right y-coordinate of the bounding box  
    area = (x2 - x1 + 1) * (y2 - y1 + 1)  
    idxs = np.argsort(y2)  
   
    # keep looping while some indexes still remain in the indexes  
    # list  
    while len(idxs) > 0:  
        # grab the last index in the indexes list and add the  
        # index value to the list of picked indexes  
        last = len(idxs) - 1  
        i = idxs[last]  
        pick.append(i)  
   
        # find the largest (x, y) coordinates for the start of  
        # the bounding box and the smallest (x, y) coordinates  
        # for the end of the bounding box  
        xx1 = np.maximum(x1[i], x1[idxs[:last]])  
        yy1 = np.maximum(y1[i], y1[idxs[:last]])  
        xx2 = np.minimum(x2[i], x2[idxs[:last]])  
        yy2 = np.minimum(y2[i], y2[idxs[:last]])  
   
        # compute the width and height of the bounding box  
        w = np.maximum(0, xx2 - xx1 + 1)  
        h = np.maximum(0, yy2 - yy1 + 1)  
   
        # compute the ratio of overlap  
        overlap = (w * h) / area[idxs[:last]]  
   
        # delete all indexes from the index list that have  
        idxs = np.delete(idxs, np.concatenate(([last],  
            np.where(overlap > overlapThresh)[0])))  
   
    # return only the bounding boxes that were picked using the  
    # integer data type  
    return boxes[pick].astype("int")

 

非极大值抑制(NMS)

标签:value   minimum   pre   numpy   nat   def   搜索   就是   oat   

原文地址:https://www.cnblogs.com/fangpengchengbupter/p/8283385.html

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