Python-创建元组矩阵

Question

我需要创建一个9x9元组矩阵。我已经有一个代码：

squares, sub = [], []
for y in [3, 6, 9]:
    sub.append([[(y, x) for x in range(1, 10) if x % 3 == 0] for x in range(1, 10) if x % 3 == 0])

for i in sub:
    sq = []
    for j in i:
        sq.extend(j)
    sq.sort()
    squares.append(sq)
    squares.append(sq)
    squares.append(sq)

具有这样的结果：

[[(3, 3), (3, 3), (3, 3), (3, 6), (3, 6), (3, 6), (3, 9), (3, 9), (3, 9)],
 [(3, 3), (3, 3), (3, 3), (3, 6), (3, 6), (3, 6), (3, 9), (3, 9), (3, 9)],
 [(3, 3), (3, 3), (3, 3), (3, 6), (3, 6), (3, 6), (3, 9), (3, 9), (3, 9)],
 [(6, 3), (6, 3), (6, 3), (6, 6), (6, 6), (6, 6), (6, 9), (6, 9), (6, 9)],
 [(6, 3), (6, 3), (6, 3), (6, 6), (6, 6), (6, 6), (6, 9), (6, 9), (6, 9)],
 [(6, 3), (6, 3), (6, 3), (6, 6), (6, 6), (6, 6), (6, 9), (6, 9), (6, 9)],
 [(9, 3), (9, 3), (9, 3), (9, 6), (9, 6), (9, 6), (9, 9), (9, 9), (9, 9)],
 [(9, 3), (9, 3), (9, 3), (9, 6), (9, 6), (9, 6), (9, 9), (9, 9), (9, 9)],
 [(9, 3), (9, 3), (9, 3), (9, 6), (9, 6), (9, 6), (9, 9), (9, 9), (9, 9)]]

结果完全符合我的要求，但是您认为有更好的书写方式吗？可以简化吗？

Answer 1

问题有点不清楚，因为最终列表（和代码）可能会被解释，并且可能有多个解释且结果相同。

对于这样的一种解释，您可以使用（扭曲的且看起来很有趣的）嵌套列表推导（[Python 3.Docs]: Data Structures - List Comprehensions：]

>>> from pprint import pprint as pp  # For print purposes only
>>>
>>> l = [3, 6, 9]
>>>
>>> items = [[i0 for i1 in l for i0 in ((i2, i1),) * len(l)] for i2 in l for _ in l]
>>>
>>> pp(items)
[[(3, 3), (3, 3), (3, 3), (3, 6), (3, 6), (3, 6), (3, 9), (3, 9), (3, 9)],
 [(3, 3), (3, 3), (3, 3), (3, 6), (3, 6), (3, 6), (3, 9), (3, 9), (3, 9)],
 [(3, 3), (3, 3), (3, 3), (3, 6), (3, 6), (3, 6), (3, 9), (3, 9), (3, 9)],
 [(6, 3), (6, 3), (6, 3), (6, 6), (6, 6), (6, 6), (6, 9), (6, 9), (6, 9)],
 [(6, 3), (6, 3), (6, 3), (6, 6), (6, 6), (6, 6), (6, 9), (6, 9), (6, 9)],
 [(6, 3), (6, 3), (6, 3), (6, 6), (6, 6), (6, 6), (6, 9), (6, 9), (6, 9)],
 [(9, 3), (9, 3), (9, 3), (9, 6), (9, 6), (9, 6), (9, 9), (9, 9), (9, 9)],
 [(9, 3), (9, 3), (9, 3), (9, 6), (9, 6), (9, 6), (9, 9), (9, 9), (9, 9)],
 [(9, 3), (9, 3), (9, 3), (9, 6), (9, 6), (9, 6), (9, 9), (9, 9), (9, 9)]]

注意：尝试修改最终列表（squares[0][0] = 1）可能会产生“意外”结果。