IoU（Intersection over Union），又称重叠度/交并比。
    1 NMS：当在图像中预测多个proposals、pred bboxes时，由于预测的结果间可能存在高冗余（即同一个目标可能被预测多个矩形框），因此可以过滤掉一些彼此间高重合度的结果；具体操作就是根据各个bbox的score降序排序，剔除与高score bbox有较高重合度的低score bbox，那么重合度的度量指标就是IoU；
    2 mAP：得到检测算法的预测结果后，需要对pred bbox与gt bbox一起评估检测算法的性能，涉及到的评估指标为mAP，那么当一个pred bbox与gt bbox的重合度较高（如IoU score > 0.5），且分类结果也正确时，就可以认为是该pred bbox预测正确，这里也同样涉及到IoU的概念；
    提到IoU，大家都知道怎么回事，讲起来也都头头是道，我拿两个图示意下（以下两张图都不是本人绘制）：
    image.png
    IOU = gt bbox、pred bbox交集的面积 / 二者并集的面积
    候选框为pre,真实框为gt
    image.png

    代码实现：

    1. from __future__ import print_function, absolute_import
    2. import numpy as np
    4. def get_IoU(pred_bbox, gt_bbox):
    6.     # bbox should be valid, actually we should add more judgements, just ignore here...
    7.     # assert ((abs(pred_bbox[2] - pred_bbox[0]) > 0) and
    8.     #         (abs(pred_bbox[3] - pred_bbox[1]) > 0))
    9.     # assert ((abs(gt_bbox[2] - gt_bbox[0]) > 0) and
    10.     #         (abs(gt_bbox[3] - gt_bbox[1]) > 0))
    12.     # -----0---- get coordinates of inters
    13.     ixmin = max(pred_bbox[0], gt_bbox[0])
    14.     iymin = max(pred_bbox[1], gt_bbox[1])
    15.     ixmax = min(pred_bbox[2], gt_bbox[2])
    16.     iymax = min(pred_bbox[3], gt_bbox[3])
    17.     iw = np.maximum(ixmax - ixmin + 1., 0.)
    18.     ih = np.maximum(iymax - iymin + 1., 0.)
    20.     # -----1----- intersection
    21.     inters = iw * ih
    23.     # -----2----- union, uni = S1 + S2 - inters
    24.     uni = ((pred_bbox[2] - pred_bbox[0] + 1.) * (pred_bbox[3] - pred_bbox[1] + 1.) +
    25.            (gt_bbox[2] - gt_bbox[0] + 1.) * (gt_bbox[3] - gt_bbox[1] + 1.) -
    26.            inters)
    28.     # -----3----- iou
    29.     overlaps = inters / uni
    31.     return overlaps
    34. def get_max_IoU(pred_bboxes, gt_bbox):
    35.     """
    36.     given 1 gt bbox, >1 pred bboxes, return max iou score for the given gt bbox and pred_bboxes
    37.     :param pred_bbox: predict bboxes coordinates, we need to find the max iou score with gt bbox for these pred bboxes
    38.     :param gt_bbox: ground truth bbox coordinate
    39.     :return: max iou score
    40.     """
    42.     # bbox should be valid, actually we should add more judgements, just ignore here...
    43.     # assert ((abs(gt_bbox[2] - gt_bbox[0]) > 0) and
    44.     #         (abs(gt_bbox[3] - gt_bbox[1]) > 0))
    46.     if pred_bboxes.shape[0] > 0:
    47.         # -----0---- get coordinates of inters, but with multiple predict bboxes
    48.         ixmin = np.maximum(pred_bboxes[:, 0], gt_bbox[0])
    49.         iymin = np.maximum(pred_bboxes[:, 1], gt_bbox[1])
    50.         ixmax = np.minimum(pred_bboxes[:, 2], gt_bbox[2])
    51.         iymax = np.minimum(pred_bboxes[:, 3], gt_bbox[3])
    52.         iw = np.maximum(ixmax - ixmin + 1., 0.)
    53.         ih = np.maximum(iymax - iymin + 1., 0.)
    55.         # -----1----- intersection
    56.         inters = iw * ih
    58.         # -----2----- union, uni = S1 + S2 - inters
    59.         uni = ((gt_bbox[2] - gt_bbox[0] + 1.) * (gt_bbox[3] - gt_bbox[1] + 1.) +
    60.                (pred_bboxes[:, 2] - pred_bboxes[:, 0] + 1.) * (pred_bboxes[:, 3] - pred_bboxes[:, 1] + 1.) -
    61.                inters)
    63.         # -----3----- iou, get max score and max iou index
    64.         overlaps = inters / uni
    65.         ovmax = np.max(overlaps)
    66.         jmax = np.argmax(overlaps)
    68.     return overlaps, ovmax, jmax
    70. if __name__ == "__main__":
    72.     # test1
    73.     pred_bbox = np.array([50, 50, 90, 100])   # top-left: <50, 50>, bottom-down: <90, 100>, <x-axis, y-axis>
    74.     gt_bbox = np.array([70, 80, 120, 150])
    75.     print (get_IoU(pred_bbox, gt_bbox))
    77.     # test2
    78.     pred_bboxes = np.array([[15, 18, 47, 60],
    79.                           [50, 50, 90, 100],
    80.                           [70, 80, 120, 145],
    81.                           [130, 160, 250, 280],
    82.                           [25.6, 66.1, 113.3, 147.8]])
    83.     gt_bbox = np.array([70, 80, 120, 150])
    84.     print (get_max_IoU(pred_bboxes, gt_bbox))