计算两个矩形的交并比，通常在检测任务里面可以作为一个检测指标。你的预测bbox和groundtruth之间的差异，就可以通过IOU来体现。很简单的算法实现，我也随便写了一个，嗯，很简单。

1. 使用时，请注意bbox四个数字的顺序(y0,x0,y1,x1)，顺序不太一样。

#!/usr/bin/env python
# encoding: utf-8



def compute_iou(rec1, rec2):
    """
    computing IoU
    :param rec1: (y0, x0, y1, x1), which reflects
            (top, left, bottom, right)
    :param rec2: (y0, x0, y1, x1)
    :return: scala value of IoU
    """
    # computing area of each rectangles
    S_rec1 = (rec1[2] - rec1[0]) * (rec1[3] - rec1[1])
    S_rec2 = (rec2[2] - rec2[0]) * (rec2[3] - rec2[1])

    # computing the sum_area
    sum_area = S_rec1 + S_rec2

    # find the each edge of intersect rectangle
    left_line = max(rec1[1], rec2[1])
    right_line = min(rec1[3], rec2[3])
    top_line = max(rec1[0], rec2[0])
    bottom_line = min(rec1[2], rec2[2])

    # judge if there is an intersect
    if left_line >= right_line or top_line >= bottom_line:
        return 0
    else:
        intersect = (right_line - left_line) * (bottom_line - top_line)
        return (intersect / (sum_area - intersect))*1.0


if __name__=='__main__':
    rect1 = (661, 27, 679, 47)
    # (top, left, bottom, right)
    rect2 = (662, 27, 682, 47)
    iou = compute_iou(rect1, rect2)
    print(iou)