deep_sort源码剖析（1）deep_sort...iou

deep_sort源码剖析（1）deep_sort...iou_matching

图像分类

最后更新 2022-05-06 20:05 阅读 1680

最后更新 2022-05-06 20:05

阅读 1680

图像分类

这是我在csdn上写的文章，现在迁移到这里

论文题目：SIMPLE ONLINE AND REALTIME TRACKING WITH A DEEP ASSOCIATION METRIC

相关博文推荐：https://blog.csdn.net/cdknight_happy/article/details/79731981

GitHub地址：https://github.com/nwojke/deep_sort

OK,话不多说正式开始

# vim: expandtab:ts=4:sw=4
from __future__ import absolute_import
import numpy as np
from . import linear_assignment

def iou(bbox, candidates):
    """Computer intersection over union.
    Parameters
    ----------
    bbox : ndarray
        A bounding box in format `(top left x, top left y, width, height)`.
    candidates : ndarray
        A matrix of candidate bounding boxes (one per row) in the same format
        as `bbox`.
    Returns
    -------
    ndarray
        The intersection over union in [0, 1] between the `bbox` and each
        candidate. A higher score means a larger fraction of the `bbox` is
        occluded by the candidate.
    """
    #预测框的左下，右上坐标
    bbox_tl, bbox_br = bbox[:2], bbox[:2] + bbox[2:]
    #多个候选框的坐标
    candidates_tl = candidates[:, :2]
    candidates_br = candidates[:, :2] + candidates[:, 2:]
    #计算中间重合部分矩形框IOU大小
    tl = np.c_[np.maximum(bbox_tl[0], candidates_tl[:, 0])[:, np.newaxis],
               np.maximum(bbox_tl[1], candidates_tl[:, 1])[:, np.newaxis]]
    br = np.c_[np.minimum(bbox_br[0], candidates_br[:, 0])[:, np.newaxis],
               np.minimum(bbox_br[1], candidates_br[:, 1])[:, np.newaxis]]
    wh = np.maximum(0., br - tl)
    #求面积
    area_intersection = wh.prod(axis=1)
    area_bbox = bbox[2:].prod()
    area_candidates = candidates[:, 2:].prod(axis=1)
    #返回IOU大小
    return area_intersection / (area_bbox + area_candidates - area_intersection)

def iou_cost(tracks, detections, track_indices=None,
             detection_indices=None):
    """An intersection over union distance metric.
    Parameters
    ----------
    tracks : List[deep_sort.track.Track]
        A list of tracks.
    detections : List[deep_sort.detection.Detection]
        A list of detections.
    track_indices : Optional[List[int]]
        A list of indices to tracks that should be matched. Defaults to
        all `tracks`.
    detection_indices : Optional[List[int]]
        A list of indices to detections that should be matched. Defaults
        to all `detections`.
    Returns
    -------
    ndarray
        Returns a cost matrix of shape
        len(track_indices), len(detection_indices) where entry (i, j) is
        `1 - iou(tracks[track_indices[i]], detections[detection_indices[j]])`.
    
    """
    #创建ID
    if track_indices is None:
        track_indices = np.arange(len(tracks))
    if detection_indices is None:
        detection_indices = np.arange(len(detections))

cost_matrix = np.zeros((len(track_indices), len(detection_indices)))
    for row, track_idx in enumerate(track_indices):
        #这个if循环我没有搞懂什么意思，知道的小伙伴可以留言告诉我
        #linear_assignment是怎么调用INFTY_COST
        #这个IF语句注释掉后，我没有看到程序有什么变化
        if tracks[track_idx].time_since_update > 1:
            cost_matrix[row, :] = linear_assignment.INFTY_COST
            continue
        计算IOU的cost值，cost越小，IOU越大
        bbox = tracks[track_idx].to_tlwh()
        candidates = np.asarray([detections[i].tlwh for i in detection_indices])
        cost_matrix[row, :] = 1. - iou(bbox, candidates)
    return cost_matrix

本作品采用知识共享署名-非商业性使用-相同方式共享 4.0 国际许可协议进行许可，转载请附上原文出处链接和本声明。
本文链接地址：https://flyai.com/article/882

讨论

500字

表情

发送

删除确认

是否删除该条评论？

取消删除