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Hi I am working on python. I created a dataframe from a csv file. One column "name" which is a text column, has inside in different places this pattern ' (some_number + %)', example:

"145 wefwignweon (100%), 1rberbebe (50%), vwrbvwrbe (100%), 140 ewggrrwrg"

I need to delete from this column where says: ' (100%)', '(100%), '(50%') In other columns are different percentage values

import pandas as pd

path_to_dir="/Users/user/Documents/file/"
name='owner.csv'
df_owner = pd.read_csv(path_to_dir+name, encoding='windows-1252') 
#df_owner["name"] =  df_owner["name"] drop where says => (' (@some_number%)')

How I can create like a kind of regular expression to drop where find this kind of values something like this? delete where says '( some_number + %)' in name column from df_owner dataframe

Regards

Webb
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2 Answers2

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You can use the regular expression \(\d+%\):

df = df[~df['name'].str.contains(r' \(\d+%\)', regex=True)]
  • OP wants the pattern removed. Your answer seems to remove the entire row of whenever the pattern is found. That is, they want `"145 wefwignweon (100%), 1rberbebe (50%), vwrbvwrbe (100%), 140 ewggrrwrg"` replaced with `"145 wefwignweon , 1rberbebe , vwrbvwrbe ), 140 ewggrrwrg"`. – Acccumulation Mar 19 '22 at 02:30
  • This showing me the columns with (100%) , (50%) – Webb Mar 19 '22 at 04:27
  • Will you please provide an actual printout of your dataframe, or a portion thereof? Thank you :) –  Mar 19 '22 at 14:08
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Capture all numbers up to three digits gives r'\d{1,3}'

But you also seem to want the parentheses, and they and the percentage sign have to be escaped, so that will be r'\(\d{1,3}\)\%'. You can then replace occurrences of that regex with the null string with lambda x: re.sub(r'\(\d{1,3}\)\%', '', x). You also might want to add the leading space to the regex.

Acccumulation
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  • Hi , i see these values: 1450 fwegwrherheheh 4556 gwregerh 4804 nifwengiwn should be 1450 fwegwrherheheh 4556 gwregerh 4804 nifwengiwn – Webb Mar 19 '22 at 04:21