Question

假设我得到一个类似于电话键盘的矩阵，

如何生成以下字典而无需亲自键入以下内容（元组不一定要排序）-

my_dict = {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)}

这本字典基本上告诉我给定数字的所有相邻数字。例如，相邻的数字1是1、2、4、5。

编辑：理想情况下，矩阵将存储为列表列表：

[[1, 2, 3], [4, 5, 6], [7, 8, 9], [None, 0, None]]

我知道暴力破解方法，但是想知道一种有效地做到这一点的方法。

Answer 1

假定小键盘存储为位置字典（元组x，y），并将相应的数字存储为值，则可以执行以下操作：

import itertools


def 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)}

这个想法是针对每个位置获取相邻点的值。

更新

要将列表的列表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。

Answer 2

您可以使用生成器功能：

import re
def 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:
       pass


s = """
1 2 3
4 5 6
7 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 re
def 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:
      pass

new_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)}

Answer 3

假设您的矩阵看起来像这样：

m = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9],
    [None, 0, None]
]

您可以蛮力解决方案，并使用defaultdict遍历矩阵并收集相邻的单元格：

from collections import defaultdict
from pprint import pprint

m = [
    [1, 2, 3],
    [4, 5, 6],
    [7, 8, 9],
    [None, 0, None]
]

rows = len(m)
cols = len(m[0])

# adjacent cells
adjacency = [(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 cells
pprint({k: tuple(sorted(v)) for k, v in d.items()})

哪个给出了排序后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)}

Answer 4

T9 = [
    [1,2,3],
    [4,5,6],
    [7,8,9],
    [None,0,None]
]

result = {T9[i][j]:[T9[i+di][j+dj] for dj in range(-1,2) for di in range(-1,2) if 0<= i+di <=3 and 0 <= j+dj <=2 and T9[i+di][j+dj] is not None]  for j in range(3) for i in range(4) if T9[i][j] is not None}

丑陋但有效

如何从给定的矩阵创建字典

4 个答案: