使用apply lambda和str从python列中获取子字符串 [英] Getting substring from column in python using apply lambda and str

问题描述

我有一个带有Fib列的数据框，我试图从中获取一个子字符串:

I have a dataframe with a column Fib, I am trying to grab a substring from it:

任何人都可以告诉我为什么此代码不起作用:

Could anyone please tell me why this code does not work:

df['new'] = df['Fib'].apply(lambda x:x.str[2:10])

AttributeError:'str'对象没有属性'str'

AttributeError: 'str' object has no attribute 'str'

但是，如果我这样做，它将起作用:

But if i do this, it will work:

df['new_col'] = df['Fib'].astype(str).str[2:10]

我正在尝试使用apply + lambda解决上述问题，只是为了获得一些经验. 谢谢

I am trying to solve the above problem with apply+lambda just to get some experience with it. Thank you

推荐答案

您的代码中的问题是，您apply沿系列行的lambda函数将收到显示的字符串.这是一个说明这一点的示例:

The problem in your code is that the lambda function you apply along the rows of your series will be receiving a string as it appears. Here's an example to illustrate this:

df = pd.DataFrame({'num':[1,4,2], 'alpha':['apple','orange','peach']})
df['alpha'].apply(lambda x:type(x))
<class 'str'>
<class 'str'>
<class 'str'>

请注意， Series.str 方法仅适用于Series，如文档中明确指出的那样:

Note that Series.str methods are only for Series, as clearly stated in the documentation:

系列和索引的向量化字符串函数

Vectorized string functions for Series and Index

因此对于您的示例，应避免使用Apply.而是:

So for your example you should avoid using apply. Instead do:

df['alpha'].str[2:10]

0     ple
1    ange
2     ach
Name: alpha, dtype: object

如果您要使用的是apply而不是您提到的那样，则只需将lambda x: x[2:10]直接切成字符串即可:

If what you want is to use apply instead as you mention, you simply need lambda x: x[2:10] as you are directly slicing the string:

df['alpha'].apply(lambda x: x[2:10])
0     ple
1    ange
2     ach
Name: alpha, dtype: object

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