def get_group_from_pair(pair_list): """ 功能:根据成对的关系,获得group 输入:对关系list,如[[1, 3], [2, 3], [4, 5], [3, 6], [6, 8]] 输出:组关系list,如[[1, 2, 3, 6, 8], [4, 5]] 方法:dfs, 先获得相邻关系,再串联 """ def get_pair_relation(pair_list): relation = {} for pair in pair_list: p1, p2 = pair relation.setdefault(p1, []).append(p2) relation.setdefault(p2, []).append(p1) return relation relation = get_pair_relation(pair_list) merged_set = set() group = [] for head in relation.keys(): if head in merged_set: continue # 已经被合并 group_item = [] que = [head] while que: item = que.pop(0) group_item.append(item) merged_set.add(item) for item_ in relation.get(item, []): if item_ not in que and item_ not in merged_set: que.append(item_) group.append(group_item) return group def nms(bbox_list, iou_thr=0.2): """ 功能:给定可能有压盖的带分数的矩形框list,输出压盖小的list 输入:[x1, y1, x2, y2, score]list, 如[[3, 8, 9, 3, 0.9], [5, 7, 8, 4, 0.7], [7, 5, 10, 1, 0.8]]; 输出:剩余的[x1, y1, x2, y2, score]list, 方法:按分数排序,被压盖的直接打掉 """ def cal_iou(bbox1, bbox2): x1, y1, x2, y2 = bbox1[:4] a1, b1, a2, b2 = bbox2[:4] p1 = min(x2, a2) - max(x1, a1) p2 = min(y2, b2) - max(y1, b1) inter = 0 if p1 > 0 and p2 > 0: inter = p1 * p2 union = (x2 - x1) * (y2 - y1) + (a2 - a1) * (b2 - b1) - inter iou = inter / union return iou bbox_list = sorted(bbox_list, key=lambda x: x[-1], reverse=True) num = len(bbox_list) reserve_flag = [True for _ in range(num)] res = [] for i in range(num): if reserve_flag[i] is False: continue # 已经被抑制 res.append(bbox_list[i]) for j in range(i + 1, num): iou = cal_iou(bbox_list[i], bbox_list[j]) if iou > iou_thr: reserve_flag[j] = False return res if __name__ == '__main__': pair_list = [[1, 3], [2, 3], [4, 5], [3, 6], [6, 8]] print(get_group_from_pair(pair_list)) bbox_list = [[3, 3, 9, 8, 0.9], [5, 4, 8, 7, 0.7], [7, 1, 10, 5, 0.8]] # bbox_list = [[5, 7, 9, 15, 0.5], [12, 8, 16, 12, 0.7], [14, 5, 20, 13, 0.9]] print(nms(bbox_list))标签:转组,group,list,item,极大值,relation,bbox,pair,配对 From: https://www.cnblogs.com/EstherLjy/p/17005850.html