快速 NMS 算法抑制没有重叠的框答案

【问题标题】：Fast NMS algorithm suppresses boxes without overlap快速 NMS 算法抑制没有重叠的框
【发布时间】：2019-05-08 02:58:27
【问题描述】：

我正在测试Fast NMS algorithm by Malisiewicz et al。我在运行示例时注意到，在一种情况下，如果我输入两个没有重叠的特定框，并且 IoU 阈值低于大约 0.75，则无论如何都会抑制一个框。

我误解了 NMS 吗？我认为如果它们之间的重叠为零，则不应丢弃任何框，无论 IoU 阈值设置在何处。

例子：

import numpy as np

def non_max_suppression_fast(boxes, overlapThresh):

    # if there are no boxes, return an empty list
    if len(boxes) == 0:
        return []

    # 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
    return boxes[pick]


# Two test boxes
                   #xmin,ymin,xmax,ymax
boxes = np.vstack([[0.3, 0.2, 0.4, 0.5], 
                  [0.1, 0.1, 0.2, 0.2]])


# no box suppression
print(non_max_suppression_fast(boxes, overlapThresh=.75))

# one box is suppressed
print(non_max_suppression_fast(boxes, overlapThresh=.74))

【问题讨论】：

标签： python algorithm object-detection non-maximum-suppression

【解决方案1】：

您的输入测试用例不合法，参数boxes 期望框坐标为绝对格式，例如像素坐标。

你可以注意到，在计算所有框的面积时，它是

area = (x2 - x1 + 1) * (y2 - y1 + 1)

+1 是添加的像素，确保 area 是框占用的实际像素数。

试试这个：

boxes = np.vstack([[3, 2, 4, 5], 
              [1, 1, 2, 2]])

【讨论】：