在Python中剥离列表 [英] Strip a list in Python

问题描述

我在python中有一个列表，它是由许多函数产生的.它是这样产生的:

I have a list in python which is produced from a number of functions. It is produced like this:

['01', '1.0', '[0.2]']

我想在生成所有不需要的字符后将其删除.

I would like to strip all of the unnecessary characters after it is produced.

理想的输出:

['01', '1.0', '0.2']

我基本上需要从列表中的最后一个字符串中删除[和].

I basically need to remove the [ and ] from the final string in the list.

更新:

list_test = ['01', '1.0', '[0.2]']
[i.strip('[]') if type(i) == str else str(i) for i in list_test]
print(list_test)

无论有没有产生相同的结果，这都不起作用:

This doesn't work as both with and without produce the same result:

['01', '1.0', '[0.2]']

所需的结果是:

['01', '1.0', '0.2']

---

提供的解决方案:

l = ['01', '1.0', '[0.2]']
[i.strip('[0.2]') if type(i) == str else str(i) for i in l]
print(l)

输出:

['01', '1.0', '0.2']

Process finished with exit code 0

推荐答案

如果你只需要删除一个可选的 [ 和 ] 围绕一个字符串，这个单行应该做到这一点:

If you just need to remove an optional [ and ] around a string, this one-liner should do the trick:

l = [i.strip('[]') for i in l]

以您的示例

>>> l = ['01', '1.0', '[0.2]']
>>> [i.strip('[]') for i in l]
['01', '1.0', '0.2']

它的作用:遍历列表的所有字符串，并删除所有 [和] (如果它们位于字符串的开头或结尾).它还会将] 0.2 [变成 0.2 .

What it does: it iterates over all strings of the list and removes any [ and ] if they are at the beginning or end of the string. It also would make ]0.2[ into 0.2.

更新:

我得到 AttributeError:'int'对象没有属性'strip'

在这种情况下，您的输入数组中有一个int，这个衬里应该可以解决问题:

In this case you have an int in your input array, this one liner should do the trick:

l = [i.strip('[]') if type(i) == str else str(i) for i in l]

以一个带有int的示例为例:

With an example with an int:

>>> l = ['01', '1.0', '[0.2]', 2]
>>> [i.strip('[]') if type(i) == str else str(i) for i in l]
['01', '1.0', '0.2', '2']

它是做什么的

• 仅从字符串中剥离 [和]
• 将非字符串的所有内容转换为字符串

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