非极大值抑制NMS

最新推荐文章于 2026-01-02 14:36:37 发布

原创最新推荐文章于 2026-01-02 14:36:37 发布 · 318 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#numpy #python

本文介绍了一个用于计算机视觉的非极大抑制（Non-Maximum Suppression，NMS）算法，它在目标检测中筛选出重叠度低于阈值的边界框，同时按置信度降序排列。通过实例和代码，展示了如何应用NMS来优化多边形检测结果并减少冗余。

输入：

边界框boxes（左上角坐标+宽高），框框间被允许的最大重叠率max_bbox_overlap，边界框置信度score

输出：

筛选后按置信度从大到小的索引pick


# -*- coding: utf-8 -*-
import numpy as np
import cv2


def non_max_suppression(boxes, max_bbox_overlap=0.6, scores=None):
    """Suppress overlapping detections.

    Original code from [1]_ has been adapted to include confidence score.

    .. [1] http://www.pyimagesearch.com/2015/02/16/
           faster-non-maximum-suppression-python/

    Examples
    --------

        >>> boxes = [d.roi for d in detections]
        >>> scores = [d.confidence for d in detections]
        >>> indices = non_max_suppression(boxes, max_bbox_overlap, scores)
        >>> detections = [detections[i] for i in indices]

    Parameters
    ----------
    boxes : ndarray
        Array of ROIs (x, y, width, height).
    max_bbox_overlap : float
        ROIs that overlap more than this values are suppressed.
    scores : Optional[array_like]
        Detector confidence score.

    Returns
    -------
    List[int]
        Returns indices of detections that have survived non-maxima suppression.

    """
    if len(boxes) == 0:
        return []

    boxes = boxes.astype(np.float)
    pick = []
    
    #tlwh to tlbr
    x1 = boxes[:, 0]
    y1 = boxes[:, 1]
    x2 = boxes[:, 2] + boxes[:, 0]
    y2 = boxes[:, 3] + boxes[:, 1]

    area = (x2 - x1 + 1) * (y2 - y1 + 1)
    if scores is not None:
        idxs = np.argsort(scores) #返回置信度从小到大的索引
    else:
        idxs = np.argsort(y2)

    while len(idxs) > 0:
        last = len(idxs) - 1
        i = idxs[last]
        pick.append(i)

        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]])

        w = np.maximum(0, xx2 - xx1 + 1)
        h = np.maximum(0, yy2 - yy1 + 1)

        #w*h为当前bbox与剩余bbox的重叠面积
        overlap = (w * h) / area[idxs[:last]]

        #删除当前框和与其高重复度框的索引
        idxs = np.delete(
            idxs, np.concatenate(
                ([last], np.where(overlap > max_bbox_overlap)[0])))

    return pick