萧十郎
假设键盘存储为位置字典（元组 x，y）和相应的数字作为值，您可以执行以下操作：import itertoolsdef distance(p1, p2): return sum((x1 - x2) ** 2 for x1, x2 in zip(p1, p2))def neighbors(positions, target): return [position for position in positions if distance(target, position) < 4]def numbers(kpad, keys): return tuple(sorted(map(kpad.get, keys)))values = list(range(1, 10)) + [0]positions = list(itertools.product([0, 1, 2], repeat=2)) + [(3, 1)]keypad = dict(zip(positions, values))result = {value: numbers(keypad, neighbors(keypad, key)) for key, value in keypad.items()}print(result)输出{0: (0, 7, 8, 9), 1: (1, 2, 4, 5), 2: (1, 2, 3, 4, 5, 6), 3: (2, 3, 5, 6), 4: (1, 2, 4, 5, 7, 8), 5: (1, 2, 3, 4, 5, 6, 7, 8, 9), 6: (2, 3, 5, 6, 8, 9), 7: (0, 4, 5, 7, 8), 8: (0, 4, 5, 6, 7, 8, 9), 9: (0, 5, 6, 8, 9)}这个想法是为每个位置获取相邻点的值。更新要将列表 a 转换为键盘字典，您可以执行以下操作：data = [[1, 2, 3], [4, 5, 6], [7, 8, 9], [None, 0, None]]keypad = {(i, j): value for i, sub in enumerate(data) for j, value in enumerate(sub) if value is not None}其余方法保持不变。
杨魅力
您可以使用生成器函数：import redef all_adjacent(_c, _graph): _funcs = [lambda x,y:(x+1, y), lambda x,y:(x+1, y+1), lambda x,y:(x+1, y-1), lambda x,y:(x, y+1), lambda x,y:(x-1, y+1), lambda x,y:(x-1, y-1), lambda x,y:(x, y-1), lambda x,y:(x-1, y)] yield _graph[_c[0]][_c[1]] for func in _funcs: a, b = func(*_c) try: if a >= 0 and b >= 0: _val = _graph[a][b] if _val != ' ': yield _val except: passs = """1 2 34 5 67 8 9 0 """new_data = [re.findall('\d+|\s{2,}', i) for i in filter(None, s.split('\n'))]final_results = {c:list(all_adjacent((i, d), new_data)) for i, a in enumerate(new_data) for d, c in enumerate(a) if c != ' '}_result = {int(a):tuple(sorted(map(int, b))) for a, b in final_results.items()}输出：{1: (1, 2, 4, 5), 2: (1, 2, 3, 4, 5, 6), 3: (2, 3, 5, 6), 4: (1, 2, 4, 5, 7, 8), 5: (1, 2, 3, 4, 5, 6, 7, 8, 9), 6: (2, 3, 5, 6, 8, 9), 7: (0, 4, 5, 7, 8), 8: (0, 4, 5, 6, 7, 8, 9), 9: (0, 5, 6, 8, 9), 0: (0, 7, 8, 9)}编辑：将矩阵存储为列表列表：import redef all_adjacent(_c, _graph): _funcs = [lambda x,y:(x+1, y), lambda x,y:(x+1, y+1), lambda x,y:(x+1, y-1), lambda x,y:(x, y+1), lambda x,y:(x-1, y+1), lambda x,y:(x-1, y-1), lambda x,y:(x, y-1), lambda x,y:(x-1, y)] yield _graph[_c[0]][_c[1]] for func in _funcs: a, b = func(*_c) try: if a >= 0 and b >= 0: _val = _graph[a][b] if _val is not None: yield _val except: passnew_data = [[1, 2, 3], [4, 5, 6], [7, 8, 9], [None, 0, None]]final_results = {c:(i, d) for i, a in enumerate(new_data) for d, c in enumerate(a) if c is not None}_result = {int(a):tuple(map(int, all_adjacent(b, new_data))) for a, b in final_results.items()}输出：{1: (1, 4, 5, 2), 2: (2, 5, 6, 4, 3, 1), 3: (3, 6, 5, 2), 4: (4, 7, 8, 5, 2, 1), 5: (5, 8, 9, 7, 6, 3, 1, 4, 2), 6: (6, 9, 8, 2, 5, 3), 7: (7, 0, 8, 5, 4), 8: (8, 0, 9, 6, 4, 7, 5), 9: (9, 0, 5, 8, 6), 0: (0, 9, 7, 8)}
冉冉说
假设您的矩阵如下所示：m = [ [1, 2, 3], [4, 5, 6], [7, 8, 9], [None, 0, None]]您可以蛮力解决您的解决方案，只需遍历矩阵并使用 defaultdict 收集相邻单元格：from collections import defaultdictfrom pprint import pprintm = [ [1, 2, 3], [4, 5, 6], [7, 8, 9], [None, 0, None]]rows = len(m)cols = len(m[0])# adjacent cellsadjacency = [(i, j) for i in (-1, 0, 1) for j in (-1, 0, 1) if not i == j == 0]d = defaultdict(list)for r in range(rows): for c in range(cols): cell = m[r][c] if cell is not None: d[cell].append(cell) for x, y in adjacency: if 0 <= r + x < rows and 0 <= c + y < cols: adjacent = m[r + x][c + y] if adjacent is not None: d[cell].append(adjacent)# print sorted adjacent cellspprint({k: tuple(sorted(v)) for k, v in d.items()})这给出了排序的相邻单元格的字典：{0: (0, 7, 8, 9), 1: (1, 2, 4, 5), 2: (1, 2, 3, 4, 5, 6), 3: (2, 3, 5, 6), 4: (1, 2, 4, 5, 7, 8), 5: (1, 2, 3, 4, 5, 6, 7, 8, 9), 6: (2, 3, 5, 6, 8, 9), 7: (0, 4, 5, 7, 8), 8: (0, 4, 5, 6, 7, 8, 9), 9: (0, 5, 6, 8, 9)}