多次元配列の並び変え方法

Question

python3を使って
例えばa=[[2,5,6],
[3,7,8],
[1,4,9],
[11,10,12]]

という配列を
a=[[1, 2, 4],
[3, 5, 7],
[6, 8, 10],
[9, 11,12]]

というように行数　列数を変えずに斜めに並べ替える方法を教えてください。

Answer 1

自分で関数を作るしかないと思います。
試しに作ってみました。
もっとスマートに書けそうな気がします。

"""
>>> a=[[2,5,6],
...    [3,7,8],
...    [1,4,9],
...    [12,10,11]]
>>> slash_sort(a)
>>> print(a)
[[1, 2, 4], [3, 5, 7], [6, 8, 10], [9, 11, 12]]
>>> b=[[2,5,6],
...    [3,7,8],
...    [1,4,9]]
>>> slash_sort(b)
>>> print(b)
[[1, 2, 4], [3, 5, 7], [6, 8, 9]]
>>> c=[[2, 5, 6, 3, 7, 8],
...    [1, 4, 9, 12, 10, 11]]
>>> slash_sort(c)
>>> print(c)
[[1, 2, 4, 6, 8, 10], [3, 5, 7, 9, 11, 12]]
"""

def slash_sort(a):
    HEIGHT = len(a)
    WIDTH = len(a[0])
    y = x = 0
    for item in sorted(sum(a, [])):
        a[y][x] = item
        y, x = y + 1, x - 1
        if x < 0:
            y, x = max(0, y - WIDTH + 1), min(y, WIDTH - 1)
        if y >= HEIGHT:
            y, x = max(0, x + y + 1 - (WIDTH - 1)), min(x + y + 1, WIDTH - 1)

テスト結果

$ python3 -m doctest -v slashsort.py
Trying:
    a=[[2,5,6],
       [3,7,8],
       [1,4,9],
       [12,10,11]]
Expecting nothing
ok
Trying:
    slash_sort(a)
Expecting nothing
ok
Trying:
    print(a)
Expecting:
    [[1, 2, 4], [3, 5, 7], [6, 8, 10], [9, 11, 12]]
ok
Trying:
    b=[[2,5,6],
       [3,7,8],
       [1,4,9]]
Expecting nothing
ok
Trying:
    slash_sort(b)
Expecting nothing
ok
Trying:
    print(b)
Expecting:
    [[1, 2, 4], [3, 5, 7], [6, 8, 9]]
ok
Trying:
    c=[[2, 5, 6, 3, 7, 8],
       [1, 4, 9, 12, 10, 11]]
Expecting nothing
ok
Trying:
    slash_sort(c)
Expecting nothing
ok
Trying:
    print(c)
Expecting:
    [[1, 2, 4, 6, 8, 10], [3, 5, 7, 9, 11, 12]]
ok
1 items had no tests:
    slashsort.slash_sort
1 items passed all tests:
   9 tests in slashsort
9 tests in 2 items.
9 passed and 0 failed.
Test passed.

Comments

[Nmonmon]

ご回答ありがとうございます。やり方の指標がわかっただけでもありがたいです！！