IoU:Intersection over Union
图片来自:https://blog.****.net/iamoldpan/article/details/78799857
#传入的是真值标签和预测标签 二维数组
def bbox_iou(bbox_a, bbox_b):
if bbox_a.shape[1] != 4 or bbox_b.shape[1] != 4:
raise IndexError
# top left 这边是计算了如图上第一幅的重叠左下角坐标值(x,y)
tl = np.maximum(bbox_a[:, None, :2], bbox_b[:, :2])
# bottom right 这边是计算了如图上第一幅的重叠左上角坐标值ymax和右下角坐标值xmax
br = np.minimum(bbox_a[:, None, 2:], bbox_b[:, 2:])
#np.prod 给定轴数值的乘积 相减就得到高和宽 然后相乘
area_i = np.prod(br - tl, axis=2) * (tl < br).all(axis=2)#重叠部分面积
area_a = np.prod(bbox_a[:, 2:] - bbox_a[:, :2], axis=1)
area_b = np.prod(bbox_b[:, 2:] - bbox_b[:, :2], axis=1)
return area_i / (area_a[:, None] + area_b - area_i)