python实现IOU计算

参考博客

import cv2
import numpy as np


def CountIOU(recA, recB):
    # 两个矩形的左上角坐标(取靠右下的)
    xA = max(recA[0], recB[0])
    yA = max(recA[1], recB[1])
    # 两个矩形的右下角坐标(去靠左上的)
    xB = min(recA[2], recA[2])
    yB = min(recA[3], recA[3])
    # 计算交集的面积(如果两个矩形没有交集则为0)
    interArea = max(0, xB - xA + 1) * max(0, yB - yA + 1)
    # 计算预测框和真实框的面积
    recA_Area = (recA[2] - recA[0] + 1) * (recA[3] - recA[1] + 1)
    recB_Area = (recB[2] - recB[0] + 1) * (recB[3] - recB[1] + 1)

    # 计算IOU
    iou = interArea / float(recA_Area + recB_Area - interArea)
    return iou


img = np.zeros((512, 512, 3))
img.fill(255)

recA = [50, 50, 99, 99]
recB = [74, 74, 124, 124]

# (图片，长方形框左上角坐标, 长方形框右下角坐标， 字体颜色，字体粗细)
cv2.rectangle(img, (recA[0], recA[1]), (recA[2], recA[3]), (0, 255, 0), 1)
cv2.rectangle(img, (recB[0], recB[1]), (recB[2], recB[3]), (0, 0, 255), 1)

IOU = CountIOU(recA, recB)
# (图片，添加的文字，左上角坐标，字体，字体大小，颜色，字体粗细)
cv2.putText(img, "IOU = %.2f" % IOU, (130, 190), cv2.FONT_HERSHEY_SIMPLEX, 0.8, (0, 0, 0), 2)

cv2.imshow("img", img)
cv2.waitKey()
cv2.destroyAllWindows()