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任务描述

计算两个多边形的IOU。可以是不同类的多边形，如一个矩形和一个三角形；也可以是两个同类的多边形。

代码

import cv2
import math
from skimage.draw import polygon
from skimage.feature import peak_local_max
import torch.nn.functional as F
import numpy as npdef polygon_IOU(polygon_1, polygon_2):"""计算两个多边形的IOU:param polygon_1: [[row1, col1], [row2, col2], ...]:param polygon_2: 同上:return:"""rr1, cc1 = polygon(polygon_2[:, 0], polygon_2[:, 1])rr2, cc2 = polygon(polygon_1[:, 0], polygon_1[:, 1])try:r_max = max(rr1.max(), rr2.max()) + 1c_max = max(cc1.max(), cc2.max()) + 1except:return 0canvas = np.zeros((r_max, c_max))canvas[rr1, cc1] += 1canvas[rr2, cc2] += 1union = np.sum(canvas > 0)if union == 0:return 0intersection = np.sum(canvas == 2)return intersection / unionif __name__ == '__main__':dx = 15# 三角形输入为三个点坐标triangle_1 = np.array([[200, 100],[180, 180],[220, 180]])triangle_2 = np.array([[200 + dx, 100],[180 + dx, 180],[220 + dx, 180]])# [row, col]rect_1 = np.array([[100, 100],[100, 200],[200, 200],[200, 100]])rect_2 = np.array([[100, 150],[100, 250],[200, 250],[200, 150]])iou = polygon_IOU(rect_1, rect_2)print('IOU={:.5f}'.format(iou))