Filter DataFrame based on partial matching string from list

Question

I have a dataframe with lots of categories. Here list of some of them

Bank 

(0827) ОСП                                  
(0283) Банк ВТБ (ПАО)                       
(0822) ОСИП_ПЕНСЫ                           
(0260) АО Тинькофф Банк                     
(0755) ПАО Совкомбанк

I want to filter dataframe based on string matching. I don't want to pass entire row name, i wanna pass something like ['Совкомбанк', 'Тинькофф']. The expecting result of this is :

(0260) АО Тинькофф Банк                     
(0755) ПАО Совкомбанк

I tried df = df[df[column_name].isin(values)] but i didn't work.

Yilun Zhang · Accepted Answer · 2021-02-11 16:22:24Z

3

.isin will check for exact match. What you are looking for is .str.contains:

match_strs =  ['Совкомбанк', 'Тинькофф']
df = df[df[column_name].str.contains("(" + "|".join(match_strs) + ")")]

You can have custom regular expressions within str.contains(...) to search for whatever you want.

answered Feb 11, 2021 at 16:22

Yilun Zhang

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4 Comments

ALollz Over a year ago

You don't even need the parentheses around the pattern for .contains, mostly that's for extract with multiple possible values. +1.

Yilun Zhang Over a year ago

@ALollz That's true, but having it won't harm, and it can be handy if the search term is a phrase rather than a single word.

ALollz Over a year ago

Well no, I don't think the parenthesis don't change anything about the pattern. But they will result in a warning: This pattern has match groups. To actually get the groups, use str.extract.

Yilun Zhang Over a year ago

@ALollz Maybe I was using it wrong :P, I knew about the warning, but it at least works for my cases.

Kenan · Accepted Answer · 2021-02-11 16:22:02Z

1

If you want to just pass the names you have to clean up the Bank column

df[df['Bank'].str.split(' ').str.get(1).isin(values)]

answered Feb 11, 2021 at 16:22

Kenan

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