CODEWITHSUNDEEP
pythonregexstringpandasdataframe
J. Dykstra
J. Dykstra

Reputation: 211

Splitting string multiple times and return the result as new DataFrame

I am trying to split a pandas column repeatedly. I want to find the string in-between two strings, indefinitely. For example, lets say i have the pandas column from the input below:

import numpy as np
import pandas as pd

data=np.array([["'abc'ad32kn'def'dfannasfl[]12a'ghi'"],
              ["'jk'adf%#d1asn'lm'dfas923231sassda"],
              ["'nop'ad&@*-0'qrs'd2&*@^#!!sda'tuv'dasdj_23'w'823a&@'xyz'adfa"]])

df = pd.DataFrame({'Practice Column': data.ravel()})

print(df)

I then, would like to split these string by opening and closing quotes '...', and then take what is inside. So, my final output would be:

enter image description here

Can someone help me out? Thanks.

Upvotes: 2

Views: 127

Answers (1)

cs95
cs95

Reputation: 402483

Let's use extractall here:

df['Practice Column'].str.extractall(r"'(.*?)'").unstack(1)[0].fillna('')

match    0    1    2  3    4
0      abc  def  ghi        
1       jk   lm             
2      nop  qrs  tuv  w  xyz

The pattern '(.*?)' finds all instances of strings within the single quotes. More info - 

'      # Match opening quote
(      # Open capture group
.*?    # Non-greedy match for anything
)      # End of capture group
'      # Match closing quote

To merge this back with df, you can either use join:

v = df.join(df['Practice Column']
    .str.extractall(r"'(.*?)'").unstack(1)[0].fillna(''))

Or, assign "Practice Column" back:

v = df['Practice Column'].str.extractall(r"'(.*?)'").unstack(1)[0].fillna('')
v.insert(0, 'Practice Column', df['Practice Column'])


print(v)

match                                    Practice Column    0    1    2  3    4
a                    'abc'ad32kn'def'dfannasfl[]12a'ghi'  abc  def  ghi        
b                     'jk'adf%#d1asn'lm'dfas923231sassda   jk   lm             
c      'nop'ad&@*-0'qrs'd2&*@^#!!sda'tuv'dasdj_23'w'8...  nop  qrs  tuv  w  xyz

Another solution with a list comprehension (for performance).

import re

p = re.compile("'(.*?)'")    
pd.DataFrame([
    p.findall(s) for s in df['Practice Column']]).fillna('')

     0    1    2  3    4
0  abc  def  ghi        
1   jk   lm             
2  nop  qrs  tuv  w  xyz

This won't work if there are NaNs, so here's a modified version of the solution above. You will need to drop the NaNs first.

pd.DataFrame([
    p.findall(s) for s in df['Practice Column'].dropna()]
).fillna('')

     0    1    2  3    4
0  abc  def  ghi        
1   jk   lm             
2  nop  qrs  tuv  w  xyz

Upvotes: 3

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