IOU批量计算

import numpy as np

box1 = np.array([[0,0,100,100],[0,0,100,100]])
box2 = np.array([[50,50,100,100],[0,0,80,80]])

def calc_iou(boxes1, boxes2):

    # calculate the left up point & right down point
    lu = np.maximum(boxes1[..., :2], boxes2[..., :2])
    rd = np.minimum(boxes1[..., 2:], boxes2[..., 2:])

    # 计算机交集面积
    intersection = np.maximum(0.0, rd - lu)
    inter_square = intersection[..., 0] * intersection[..., 1]

    # 计算每个box的面积
    square1 = (boxes1[..., 2] - boxes1[..., 0])* (boxes1[..., 3] -  boxes1[..., 1])
    square2 = (boxes2[..., 2] - boxes2[..., 0])* (boxes2[..., 3] -  boxes2[..., 1])
    #计算并集面积
    union_square = np.maximum(square1 + square2 - inter_square, 1e-10)

    #numpy 后面两个参数是分别表示最大值与最小值
    return np.clip(inter_square / union_square, 0.0, 1.0)

iou  = calc_iou(box1,box2)
print(iou)