class TreeNode:    def __init__(self, x):        self.val = x        self.left = None        self.right = Noneimport numpy as np# 输入： dets为候选框; thresh为iou阈值# 返回值 keepdef nms(dets, thresh):    x1 = dets[:, 0]    y1 = dets[:, 1]    x2 = dets[:, 2]    y2 = dets[:, 3]    scores = dets[:, -1]    order = np.argsort(scores)[::-1] # 排序后的索引,由大到小    areas = (x2-x1) * (y2-y1)    # print(areas)    # 最后保留的结果    keep = []    while order.size > 0:        i = order[0] # 取出置信度最高的窗口        keep.append(i)        # 计算窗口i 和其他所有窗口 交叠部分面积        xx1 = np.maximum(x1[i], x1[order[1:]])        yy1 = np.maximum(y1[i], y1[order[1:]])        xx2 = np.minimum(x2[i], x2[order[1:]])        yy2 = np.minimum(y2[i], y2[order[1:]])        w = np.maximum(0.0, xx2 - xx1)        h = np.maximum(0.0, yy2 - yy1)        inter = w * h        # 计算交并比iou        ovr = inter / ( areas[i] + areas[order[1:] ] - inter )        print(ovr)        # inds 为 所有与窗口i的iou值小于threshold值的索引        # np.where返回值为tuple，*[0]才为真正想要的数据        inds = np.where(ovr <= thresh)[0]        print("IOU合并 ", i, order[np.where(ovr > thresh)[0]+1])        # 因为窗口i代表自身，inds + 1即可对应于order中那些不满足条件的窗口索引位置        order = order[inds + 1]    return dets[keep]import cv2 as cvimport numpy as npimport nmsboxes = [[200, 200, 400, 400, 0.99],         [220, 220, 420, 420, 0.9],         [100, 100, 150, 150, 0.82],         [200, 240, 400, 440, 0.5],         [150, 250, 300, 400, 0.88]]boxes = np.array(boxes)overlap = 0.6 # 阈值pick = nms.nms(boxes, overlap)print(pick)###canvas = np.zeros((500, 500, 3), dtype='uint8')color = (0, 255, 255)for box in boxes:    x1, y1, x2, y2, _ = map(int, box)    cv.rectangle(canvas, (x1, y1), (x2, y2), color, 5)color = (255, 0, 255)for box in pick:    # print(box)    x1, y1, x2, y2, _ = map(int, box)    cv.rectangle(canvas, (x1, y1), (x2, y2), color, 2)cv.imwrite("result.png", canvas)cv.imshow('nms', canvas)cv.waitKey()