一、交并比IOU(Intersection-over-Union,IoU)

  • 交并比
    交并比及非最大值抑制
    交并比越大,两个包围盒(bounding boxes)的重叠程度越高
  • python实现
def iou(box1, box2):
    """Implement the intersection over union (IoU) between box1 and box2
    
    Arguments:
    box1 -- first box, list object with coordinates (x1, y1, x2, y2)
    box2 -- second box, list object with coordinates (x1, y1, x2, y2)
    """

    # Calculate the (y1, x1, y2, x2) coordinates of the intersection of box1 and box2. Calculate its Area
    xi1 = max(box1[0], box2[0])
    yi1 = max(box1[1], box2[1])
    xi2 = min(box1[2], box2[2])
    yi2 = min(box1[3], box2[3])
    inter_area = max(0,(yi2-yi1))*max(0,(xi2-xi1))

    # Calculate the Union area by using Formula: Union(A,B) = A + B - Inter(A,B)
    box1_area = (box1[3]-box1[1])*(box1[2]-box1[0])
    box2_area = (box2[3]-box2[1])*(box2[2]-box2[0])
    union_area = box1_area + box2_area - inter_area
    
    # compute the IoU
    iou = inter_area/union_area

    return iou

二、 非最大值抑制(Non-Maximum Suppression,NMS)

  • NMS在边缘检测、人脸检测、目标检测(YOLO,SSD,Faster R-CNN)等计算机视觉视觉任务中得到了广泛应用

  • 为什么要非最大值抑制?目标检测的过程中在同一目标的位置上会产生大量的候选框,这些候选框相互之间可能会有重叠,故而存在模型可能对同一个对象做出多次检测的状况。此时我们需要利用非极大值抑制找到最佳的目标边界框,消除冗余的边界框。
    交并比及非最大值抑制

  • 实现方法
    预处理:score threshold (抛弃所有预测概率Pc小于一定阈值的输出边界框)
    NMS: 对于Bounding Box的列表B及其对应的置信度S,采用下面的计算方式:选择具有最大score的检测框M,将其从B集合中移除并加入到最终的检测结果D中.通常将B中剩余检测框中与M的IoU大于阈值iou_threshold的框从B中移除.重复这个过程,直到B为空。

  • tensorflow中可以使用tf.image.non_max_suppression(boxes, scores, max_boxes_tensor, iou_threshold)函数

NMS Python实现 - nms.py

# --------------------------------------------------------
# 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 nms(dets, thresh):
    x1 = dets[:, 0]
    y1 = dets[:, 1]
    x2 = dets[:, 2]
    y2 = dets[:, 3]
    scores = dets[:, 4]

    areas = (x2 - x1 + 1) * (y2 - y1 + 1)  # 各 box 的面积
    order = scores.argsort()[::-1]  #  将score降序排序,并返回排序后的下标 

    keep = []
    while order.size > 0:
        i = order[0]  #   score 最大的 box 对应的 index
        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  # 交面积
        ovr = inter / (areas[i] + areas[order[1:]] - inter)  # 交并比

        inds = np.where(ovr <= thresh)[0]  # 保留 IoU 小于设定阈值的 boxes
        order = order[inds + 1]

    return keep

python 中的[:-1]和[::-1] b=a[:-1] #从位置0到位置-1之前的数 b=a[::-1] 为倒序

相关文章: