能不能用列表推导式将二维列表的次对角线的值赋值给一个新的列表？
>>> from random import randint as rnd
>>> a = [[rnd(0,9) for _ in range(6)] for _ in range(6)]
>>> for i in a:
print(i)
[2, 0, 7, 0, 5, 4]
[0, 2, 3, 4, 9, 8]
[4, 6, 8, 8, 6, 6]
[8, 1, 4, 8, 5, 6]
[6, 7, 7, 5, 9, 7]
[5, 6, 1, 2, 6, 1]
>>> [n[-i-1] for i,n in enumerate(a) if i<len(a[0])]
[4, 9, 8, 4, 7, 5]
>>> a = [[rnd(0,9) for _ in range(6)] for _ in range(4)]
>>> for i in a:
print(i)
[0, 1, 0, 7, 1, 2]
[0, 2, 2, 9, 3, 1]
[7, 0, 5, 3, 7, 0]
[1, 7, 6, 2, 0, 5]
>>> [n[-i-1] for i,n in enumerate(a) if i<len(a[0])]
[2, 3, 3, 6]
>>> a = [[rnd(0,9) for _ in range(4)] for _ in range(6)]
>>>
>>> for i in a:
print(i)
[7, 5, 7, 1]
[5, 8, 0, 4]
[8, 6, 6, 8]
[0, 1, 7, 4]
[9, 5, 7, 7]
[2, 1, 2, 0]
>>> [n[-i-1] for i,n in enumerate(a) if i<len(a[0])]
[1, 0, 6, 0]
如不加 if i<len(a[0]) 当行数大于列数时会报错！