将所有字符串切成列表? [英] Slice all strings in a list?

问题描述

是否有Python方式将列表中的所有字符串切片?

Is there a Pythonic way to slice all strings in a list?

假设我有一个字符串列表:

Suppose I have a list of strings:

list = ['foo', 'bar', 'baz']

我只想要每个字符串的最后2个字符:

And I want just the last 2 characters from each string:

list2 = ['oo', 'ar', 'az']

我该怎么办?

我知道我可以遍历列表并从每个列表中提取list[i][-2:]，但这似乎不是Python风格的.

I know I can iterate thru the list and take list[i][-2:] from each one, but that doesn't seem very Pythonic.

一般来说，我的代码是:

Less generally, my code is:

def parseIt(filename):
    with open(filename) as f:
        lines = f.readlines()

   result = [i.split(',') for i in lines[]]

...除了我只想从每行(而不是整行)中拆分lines[i][20:].

...except I only want to split lines[i][20:] from each line (not the whole line).

推荐答案

您提到您可以按照规范进行list[i][-2:]转换列表，但是您真正要寻找的是:

You mentioned that you can do list[i][-2:] for transforming the list per your specification, but what you are actually looking for is:

[word[1:] for word in lst]

此外，对于您提供的代码示例，您希望从一开始就将其分割为20个字符，解决方案将是相同的:

Furthermore, for the code sample you provided where you are looking to slice 20 characters from the beginning, the solution would be the same:

result = [i[20:].split(',') for i in lines]

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