Question

我有一个文本文件，其数据结构如下：

01/May/1998:15:28:53    test123 0   383L    281L    399
01/May/1998:14:23:28    doe821  62C 621L    379
01/May/1998:22:10:11    testABC 0   635R    407R    671R    671N    407N    407Q    407L    496L    569

每个数据都以日期和时间开头，格式如下：01/May/1998:15:28:53。

我开始阅读文本文件，但现在我想将其转换为列表。我怎样才能做到这一点？我需要正则表达式吗？

非常感谢任何帮助。

编辑：我想要这个输出：

    [
      ['01/May/1998:15:28:53', 'test123', '0', '383L', '281L', '399'],
      ['01/May/1998:14:23:28', 'doe821', '62C', '621L', '379'],
      ['01/May/1998:22:10:11', 'testABC', '0', '635R', '407R', '671R', '671N', '407N', '407Q', '407L', '496L', '569']
    ]

Answer 1

在每一行上拨打str.split()会给您：

 ['01/May/1998:15:28:53', 'test123', '0', '383L', '281L', '399']

如：

with open('textfile') as f:
    for line in f:
        print line.split()

['01/May/1998:15:28:53', 'test123', '0', '383L', '281L', '399']
['01/May/1998:14:23:28', 'doe821', '62C', '621L', '379']
['01/May/1998:22:10:11', 'testABC', '0', '635R', '407R', '671R', '671N', '407N', '407Q', '407L', '496L', '569']

将每一行作为一个列表项：

with open('textfile') as f:
    print f.readlines() # note the newline chars(\n) that may need slicing off

['01/May/1998:15:28:53    test123 0   383L    281L    399\n', '01/May/1998:14:23:28    doe821  62C 621L    379\n', '01/May/1998:22:10:11    testABC 0   635R    407R    671R    671N    407N    407Q    407L    496L    569\n']

要将每一行拆分并放在一个大清单中：

with open('textfile') as f:
    print [line.split() for line in f]

[['01/May/1998:15:28:53', 'test123', '0', '383L', '281L', '399'], ['01/May/1998:14:23:28', 'doe821', '62C', '621L', '379'], ['01/May/1998:22:10:11', 'testABC', '0', '635R', '407R', '671R', '671N', '407N', '407Q', '407L', '496L', '569']]

Answer 2

假设您的文件名为test.data

>>> with open('test.data') as f:
>>>     [x for x in [y.split() for y in f.read().split('\n')]]

将String格式的数据转换为List - python

2 个答案: