如何使用模块re从数据框的列中删除特殊字符? [英] How to remove special characers from a column of dataframe using module re?

问题描述

嘿，我看到了该链接，但是他们没有在任何地方使用re模块，这就是我在此处发布的原因.希望您理解并删除重复项.

Hey I have seen that link but nowhere there they have used re module that's why I have posted here. Hope you understand and remove the duplicate.

这是链接.我想使用re模块.

表格:

A    B    C    D
1    Q!   W@   2
2    1$   E%   3
3    S2#  D!   4

在这里，我要删除column B和C中的特殊字符.我已经使用过.transform()，但是如果可能的话，我想使用re进行操作，但是出现错误.

here I want to remove the special characters from column B and C. I have used .transform() but I want to do it using re if possible but I am getting errors.

输出:

A    B    C    D   E   F
1    Q!   W@   2   Q   W
2    1$   E%   3   1   E
3    S2#  D!   4   S2  D

我的代码:

df['E'] = df['B'].str.translate(None, ",!.; -@!%^&*)(")

只有知道什么是特殊字符，它才有效.

It's working only if I know what are the special characters.

但是我想使用re这是最好的方法.

But I want to use re which would be the best way.

import re
#re.sub(r'\W+', '', your_string)
df['E'] = re.sub(r'\W+', '', df['B'].str)

我在这里遇到错误:

TypeError: expected string or buffer

所以我应该如何传递值以获得正确的输出.

So how should I pass the value to get the correct output.

推荐答案

作为

As this answer shows, you can use map() with a lambda function that will assemble and return any expression you like:

df['E'] = df['B'].map(lambda x: re.sub(r'\W+', '', x))

lambda仅定义匿名函数.您可以将它们保留为匿名状态，或像其他任何对象一样将它们分配给引用. my_function = lambda x: x.my_method(3)等同于def my_function(x): return x.my_method(3).

lambda simply defines anonymous functions. You can leave them anonymous, or assign them to a reference like any other object. my_function = lambda x: x.my_method(3) is equivalent to def my_function(x): return x.my_method(3).

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