如何对列表列表的第 n 列进行切片？
我想对列表列表的第 n 列进行切片。 例如，
matrix = [
["c","b","a","c"],
["d","a","f","d"],
["g","h","i","a"]
]
我需要转置列表然后使用索引吗？例如
transposed = []
for i in range(len(matrix)+1):
transposed.append([row[i] for row in matrix])
transposed[1]
>>> ['b','a','h']
，或者有没有办法直接在嵌套列表上使用索引？
我正在尝试类似的事情：
matrix[:][1]
>>>['d','a','f','d']
但这不起作用，正如我发现的那样。
谢谢
I want to slice the the nth column of a list of lists.
e.g.
matrix = [
["c","b","a","c"],
["d","a","f","d"],
["g","h","i","a"]
]
Do i need to transpose the list and then use indices? e.g.
transposed = []
for i in range(len(matrix)+1):
transposed.append([row[i] for row in matrix])
transposed[1]
>>> ['b','a','h']
or is there a way to use indices directly on the nested list?
I was trying something like:
matrix[:][1]
>>>['d','a','f','d']
but this does not work as I found out.
Thank you
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这是你的意思吗？
Is this what you meant?
您可以使用
zip转置矩阵：输出：
('b', 'a', 'h')You could use
zipto transpose your matrix:Output:
('b', 'a', 'h')