在 Python 中转置矩阵 [英] Transpose a matrix in Python

问题描述

我正在尝试用 Python 创建一个矩阵转置函数.矩阵是一个二维数组，表示为整数列表的列表.例如下面是一个2X3的矩阵(即矩阵的高为2，宽为3):

A=[[1, 2, 3],[4, 5, 6]]

要转置的第i个索引中的第j个项目应该成为第j个索引中的第i个项目.以下是上述示例的转置方式:

<预><代码>>>>转置([[1, 2, 3],[4, 5, 6]])[[1, 4],[2, 5],[3, 6]]>>>转置([[1, 2],[3, 4]])[[1, 3],[2, 4]]

我该怎么做?

解决方案

您可以使用 zip 和 * 得到矩阵的转置:

<预><代码>>>>A = [[ 1, 2, 3],[ 4, 5, 6]]>>>拉链(*A)[(1, 4), (2, 5), (3, 6)]>>>lis = [[1,2,3],... [4,5,6]，... [7,8,9]]>>>zip(*lis)[(1, 4, 7), (2, 5, 8), (3, 6, 9)]

如果您希望返回的列表是列表列表:

<预><代码>>>>[list(x) for x in zip(*lis)][[1, 4, 7], [2, 5, 8], [3, 6, 9]]#或者>>>地图(列表，zip(*lis))[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

I'm trying to create a matrix transpose function in Python. A matrix is a two dimensional array, represented as a list of lists of integers. For example, the following is a 2X3 matrix (meaning the height of the matrix is 2 and the width is 3):

A=[[1, 2, 3],
   [4, 5, 6]]

To be transposed the jth item in the ith index should become the ith item in the jth index. Here's how the above sample would look transposed:

>>> transpose([[1, 2, 3],
               [4, 5, 6]])
[[1, 4],
[2, 5],
[3, 6]]
>>> transpose([[1, 2],
               [3, 4]])
[[1, 3],
[2, 4]]

How can I do this?

解决方案

You can use zip with * to get transpose of a matrix:

>>> A = [[ 1, 2, 3],[ 4, 5, 6]]
>>> zip(*A)
[(1, 4), (2, 5), (3, 6)]
>>> lis  = [[1,2,3], 
... [4,5,6],
... [7,8,9]]
>>> zip(*lis)
[(1, 4, 7), (2, 5, 8), (3, 6, 9)]

If you want the returned list to be a list of lists:

>>> [list(x) for x in zip(*lis)]
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]
#or
>>> map(list, zip(*lis))
[[1, 4, 7], [2, 5, 8], [3, 6, 9]]

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