以下是一个简单的实现：

def match_points(points1, points2, n, error): """ 匹配两组点，当满足连续N个点的高程误差在一定范围内，且点名也相同时，匹配成功。 否则，只匹配高程，当满足连续N个点的高程误差在一定范围内时，认为匹配对应的点名是同一个点。 当匹配结束时，还有未匹配成功的点时，倒序匹配一次，仍有未匹配是，将N-1，继续匹配。 :param points1: 第一组点列表，每个元素是一个字典，包含点名和高程两个属性 :param points2: 第二组点列表，每个元素是一个字典，包含点名和高程两个属性 :param n: 连续匹配的点数 :param error: 高程误差范围 :return: 匹配结果，是一个元组列表，每个元组包含两个字典，分别表示匹配的两个点 """ result = [] matched1 = set() matched2 = set() for i in range(len(points1)): for j in range(len(points2)): if j in matched2: continue if points1[i]['name'] == points2[j]['name']: if i in matched1: break k = 1 while i + k < len(points1) and j + k < len(points2) and k < n: if abs(points1[i + k]['elevation'] - points2[j + k]['elevation']) > error: break if points1[i + k]['name'] != points2[j + k]['name']: break k += 1 if k == n: result.append((points1[i], points2[j])) matched1.add(i) matched2.add(j) break elif abs(points1[i]['elevation'] - points2[j]['elevation']) <= error: if i in matched1: break k = 1 while i + k < len(points1) and j + k < len(points2) and k < n: if abs(points1[i + k]['elevation'] - points2[j + k]['elevation']) > error: break k += 1 if k == n: result.append((points1[i], points2[j])) matched1.add(i) matched2.add(j) break if len(matched1) < len(points1) or len(matched2) < len(points2): if n > 1: return match_points(points1[::-1], points2[::-1], n - 1, error)[::-1] return result

示例用法

points1 = [{'name': 'A', 'elevation': 100}, {'name': 'B', 'elevation': 110}, {'name': 'C', 'elevation': 120}, {'name': 'D', 'elevation': 130}, {'name': 'E', 'elevation': 140}] points2 = [{'name': 'A', 'elevation': 95}, {'name': 'X', 'elevation': 115}, {'name': 'B', 'elevation': 105}, {'name': 'C', 'elevation': 125}, {'name': 'D', 'elevation': 130}, {'name': 'E', 'elevation': 145}] result = match_points(points1, points2, 3, 5) for r in result: print(r