快速NMS算法可抑制没有重叠的盒子

时间:2019-05-08 02:58:27

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

我正在测试Fast NMS algorithm by Malisiewicz et al。我在遍历示例时注意到,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0.75,则无论如何,一个框都会被抑制。

我误解了NMS吗?我认为,无论IoU阈值设置在什么位置,如果它们之间的重叠为零,则不要丢弃任何框。

示例:

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))

1 个答案:

答案 0 :(得分:1)

您的输入测试用例不合法,参数boxes期望框以绝对格式进行坐标处理,例如以像素坐标表示。

您会注意到,在计算所有框的面积时,它是

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

+1是一个增加的像素,请确保area是该框占用的实际像素数。

尝试一下:

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