算法的作用
当算法对一个目标产生了多个候选框的时候，选择 score 最高的框，并抑制其他对于改目标的候选框。
适用场景
一幅图中有多个目标（如果只有一个目标，那么直接取 score 最高的候选框即可）。
算法的输入
算法对一幅图产生的所有的候选框，以及每个框对应的 score （可以用一个 5 维数组 dets 表示，前 4 维表示四个角的坐标，第 5 维表示分数），阈值 thresh。
算法的输出
正确的候选框组（dets 的一个子集）。
细节
- 起始，设所有的框都没有被抑制，所有框按照
score从大到小排序。- 从第 0 个框（分数最高）开始遍历：对于每一个框，如果该框没有被抑制，就将所有与它
IoU大于thresh的框设为抑制。- 返回没被抑制的框。
NMS流程：
实现1：
#coding:utf-8
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] #bbox打分
areas = (x2 - x1 + 1) * (y2 - y1 + 1)
#打分从大到小排列，取index
order = scores.argsort()[::-1]
#keep为最后保留的边框
keep = []
while order.size > 0:
#order[0]是当前分数最大的窗口，肯定保留
i = order[0]
keep.append(i)
#计算窗口i与其他所有窗口的交叠部分的面积
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
#交/并得到iou值
ovr = inter / (areas[i] + areas[order[1:]] - inter)
#inds为所有与窗口i的iou值<=threshold值的窗口的index，其他窗口也就是iou>threshold的值因为重复了所以被删除，仅仅保留iou<=threshold
inds = np.where(ovr <= thresh)[0]
#order里面只保留与窗口i交叠面积小于threshold的那些窗口，由于ovr长度比order长度少1(不包含i)，所以inds+1对应到保留的窗口
order = order[inds + 1]
return keep实现2：
# import the necessary packages
import numpy as np
def non_max_suppression(boxes, probs=None, overlapThresh=0.3):
# if there are no boxes, return an empty list
if len(boxes) == 0:
return []
# if the bounding boxes are 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 grab the indexes to sort
# (in the case that no probabilities are provided, simply sort on the
# bottom-left y-coordinate)
area = (x2 - x1 + 1) * (y2 - y1 + 1)
idxs = y2
# if probabilities are provided, sort on them instead
if probs is not None:
idxs = probs
# sort the indexes
idxs = np.argsort(idxs)
# 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 overlap greater
# than the provided overlap threshold
idxs = np.delete(idxs, np.concatenate(([last],
np.where(overlap > overlapThresh)[0])))
# return only the bounding boxes that were picked
return boxes[pick].astype("int")参考：