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)
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编写方法吗？能做得更简单吗？
结果应始终相同，没有任何动态输入，以后也将不对其进行编辑。只是我的代码不同部分中使用的静态矩阵。
这个问题还不清楚，因为最终列表（和代码）可能会被解释，并且可能有多个解释具有相同的结果。
对于一种这样的解释，您可以使用（扭曲的，看起来很有趣的）嵌套列表推导（[Python 3.Docs]：数据结构-列表推导）：
>>> 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：尝试修改您的最终列表（squares[0][0] = 1）可能会产生“意外”结果。
本文收集自互联网，转载请注明来源。
如有侵权，请联系 [email protected] 删除。
我来说两句