目标检测3: Detection基础之IoU

def bb_intersection_over_union(boxA, boxB):
  # determine the (x, y)-coordinates of the intersection rectangle
  xA = max(boxA[0], boxB[0])
  yA = max(boxA[1], boxB[1])
  xB = min(boxA[2], boxB[2])
  yB = min(boxA[3], boxB[3])

  # compute the area of intersection rectangle
  interArea = max(0, xB - xA + 1) * max(0, yB - yA + 1)

  # compute the area of both the prediction and ground-truth
  # rectangles
  boxAArea = (boxA[2] - boxA[0] + 1) * (boxA[3] - boxA[1] + 1)
  boxBArea = (boxB[2] - boxB[0] + 1) * (boxB[3] - boxB[1] + 1)

  # compute the intersection over union by taking the intersection
  # area and dividing it by the sum of prediction + ground-truth
  # areas - the interesection area
  iou = interArea / float(boxAArea + boxBArea - interArea)

  # return the intersection over union value
  return iou

函数需要两个参数：boxA和boxB，假定是我们的ground-truth和预测框。第3-6行确定两个矩形框的（x，y）坐标，然后我们用它们来计算交集的面积（第9行）。interArea变量表示IoU公式中的分子。为了计算分母，我们首先需要计算出预测框和ground-truth的区域（第13和14行）。然后可以在19行上通过将交集区域除以两个边界框的并集区域与交叉区域的差来计算IoU，（注意从分母中减去交叉区域，否则交叉区域将被双倍计数）。最后，IoU分数将返回到第22行的调用函数。现在我们的IoU方法已经完成。

import numpy as np


def bbox_overlaps(bboxes1, bboxes2, mode='iou'):
    """Calculate the ious between each bbox of bboxes1 and bboxes2.
    Args:
        bboxes1(ndarray): shape (n, 4)
        bboxes2(ndarray): shape (k, 4)
        mode(str): iou (intersection over union) or iof (intersection
            over foreground)
    Returns:
        ious(ndarray): shape (n, k)
    """

    assert mode in ['iou', 'iof']

    bboxes1 = bboxes1.astype(np.float32)
    bboxes2 = bboxes2.astype(np.float32)
    rows = bboxes1.shape[0]
    cols = bboxes2.shape[0]
    ious = np.zeros((rows, cols), dtype=np.float32)
    if rows * cols == 0:
        return ious
    exchange = False
    if bboxes1.shape[0] > bboxes2.shape[0]:
        bboxes1, bboxes2 = bboxes2, bboxes1
        ious = np.zeros((cols, rows), dtype=np.float32)
        exchange = True
    area1 = (bboxes1[:, 2] - bboxes1[:, 0] + 1) * (
        bboxes1[:, 3] - bboxes1[:, 1] + 1)
    area2 = (bboxes2[:, 2] - bboxes2[:, 0] + 1) * (
        bboxes2[:, 3] - bboxes2[:, 1] + 1)
    for i in range(bboxes1.shape[0]):
        x_start = np.maximum(bboxes1[i, 0], bboxes2[:, 0])
        y_start = np.maximum(bboxes1[i, 1], bboxes2[:, 1])
        x_end = np.minimum(bboxes1[i, 2], bboxes2[:, 2])
        y_end = np.minimum(bboxes1[i, 3], bboxes2[:, 3])
        overlap = np.maximum(x_end - x_start + 1, 0) * np.maximum(
            y_end - y_start + 1, 0)
        if mode == 'iou':
            union = area1[i] + area2 - overlap
        else:
            union = area1[i] if not exchange else area2
        ious[i, :] = overlap / union
    if exchange:
        ious = ious.T
    return ious