CODEWITHSUNDEEP
pythonregexpandastextreplace
SFC
SFC

Reputation: 793

Using regex to alter digits pandas

Background

I have the following df

import pandas as pd
df = pd.DataFrame({'Text' : ['But the here is \nBase ID: 666666    \nDate is Here 123456 ', 
                                   '999998 For \nBase ID: 123456    \nDate  there', 
                                   'So so \nBase ID: 939393    \nDate hey the 123455 ',],
                      'ID': [1,2,3],
                       'P_ID': ['A','B','C'],

                     })

Output

    ID  P_ID    Text
0   1   A   But the here is \nBase ID: 666666 \nDate is Here 123456
1   2   B   999998 For \nBase ID: 123456 \nDate there
2   3   C   So so \nBase ID: 939393 \nDate hey the 123455

Tried

I have tried the following to **BLOCK** the 6 digits in between \nBase ID: and \nDate

df['New_Text'] = df['Text'].str.replace('ID:(.+?)','ID:**BLOCK**')

And I get the following 

  ID P_ID Text New_Text
0               But the here is \nBase ID:**BLOCK**666666 \nDate is Here 123456
1               999998 For \nBase ID:**BLOCK**123456 \nDate there
2               So so \nBase ID:**BLOCK**939393 \nDate hey the 123455

But I don't quite get what I want

Desired Output

  ID P_ID Text New_Text
0               But the here is \nBase ID:**BLOCK** \nDate is Here 123456
1               999998 For \nBase ID:**BLOCK** \nDate there
2               So so \nBase ID:**BLOCK** \nDate hey the 123455

Question

How do I tweak str.replace('ID:(.+?)','ID:**BLOCK**') part of my code to get my desired output?

Upvotes: 1

Views: 80

Answers (3)

noufel13
noufel13

Reputation: 663

You can try with below piece of code to get your desired output,

df['New_Text'] = df['Text'].str.replace('ID:\s+[0-9]+','ID:**BLOCK**')

Output:

0    But the here is \nCase ID:**BLOCK**    \nDate is Here 123456 
1    999998 For \nCase ID:**BLOCK**    \nDate  there              
2    So so \nCase ID:**BLOCK**    \nDate hey the 123455

Regex Breakdown:

'\s+' - to indicate space(s)

'[0-9]+' - to specify a number

Upvotes: 1

Moshel
Moshel

Reputation: 420

try df['New_Text'] = df['Text'].str.replace('ID:(.+?)\n','ID:**BLOCK**\n')

regexp match the shortest possible string, in your case ' '

Upvotes: 1

Xukrao
Xukrao

Reputation: 8633

df['New_Text'] = df['Text'].str.replace(r'ID: *\d+ *', 'ID:**BLOCK** ')

See here for a detailed break-down of the used regex pattern.

Upvotes: 1

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