python dataframe lowercase slicing

Question

I have a dataframe. I want to slice it by checking if the value contains a string. For example, this code works:

data_df[data_df['column1'].str.contains('test')]

But I first want to set my column1 to be all lowercase first. So being the n00b that I am, I tried:

data_df[data_df['column1'].lower().str.contains('test')]

Of course the Python gods gave me no mercy and gave me an AttributeError. Any tips on how I can slice a dataframe based on a substring but first make everything into lowercase first?

I feel like the following post is very close to my answer but I can't get it to work exactly how I described up there: Python pandas dataframe slicing, with if condition

Thanks Python pros!!!

You need str accessor before lower() as well: data_df[data_df['column1'].str.lower().str.contains('test')] — user2285236
– user2285236, Commented May 10, 2016 at 4:49

RSHAP · Accepted Answer · 2016-05-10 00:27:42Z

1

Try using apply()

data_df[data_df['column1'].apply(str.lower).str.contains('test')]

answered May 10, 2016 at 0:27

RSHAP

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Simon Chemnitz-Thomsen · Accepted Answer · 2022-07-01 09:43:56Z

0

You can drop the apply:

data_df[data_df['column1'].str.lower().str.contains('test')]

answered Jul 1, 2022 at 9:43

Simon Chemnitz-Thomsen

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