python - 按列读取文本文件并存储在 python 的列表中

Question

我有一个以下格式的文本文件:

a,b,c,d,
1,1,2,3,
4,5,6,7,
1,2,5,7,
6,9,8,5,

我如何有效地将其读入列表以获得以下内容输出？

list=[[1,4,1,6],[1,5,2,9],[2,6,5,8],[3,7,7,5]]

Answer 1

假设该文件名为 spam.txt :

$ cat spam.txt
a,b,c,d,
1,1,2,3,
4,5,6,7,
1,2,5,7,
6,9,8,5,    

使用list comprehensions和 zip()内置函数，您可以编写如下程序:

>>> with open('spam.txt', 'r') as file:
...     file.readline() # skip the first line
...     rows = [[int(x) for x in line.split(',')[:-1]] for line in file]
...     cols = [list(col) for col in zip(*rows)]
... 
'a,b,c,d,\n'
>>> rows
[[1, 1, 2, 3], [4, 5, 6, 7], [1, 2, 5, 7], [6, 9, 8, 5]]
>>> cols
[[1, 4, 1, 6], [1, 5, 2, 9], [2, 6, 5, 8], [3, 7, 7, 5]]

此外，zip(*rows)基于unpacking argument lists ，它解压列表或元组，以便其元素可以作为单独的位置参数传递给函数。换句话说，zip(*rows)减少为zip([1, 1, 2, 3], [4, 5, 6, 7], [1, 2, 5, 7], [6, 9, 8, 5]) .

编辑:

这是基于NumPy的版本，供引用:

>>> import numpy as np
>>> with open('spam.txt', 'r') as file:
...     ncols = len(file.readline().split(',')) - 1
...     data = np.fromiter((int(v) for line in file for v in line.split(',')[:-1]), int, count=-1)
...     cols = data.reshape(data.size / ncols, ncols).transpose()
...
>>> cols
array([[1, 4, 1, 6],
       [1, 5, 2, 9],
       [2, 6, 5, 8],
       [3, 7, 7, 5]])