Remove and replace multiple commas in string

Question

I have this dataset

df = pd.DataFrame({'name':{0: 'John,Smith', 1: 'Peter,Blue', 2:'Larry,One,Stacy,Orange' , 3:'Joe,Good' , 4:'Pete,High,Anne,Green'}})

yielding:

name
0   John,Smith
1   Peter,Blue
2   Larry,One,Stacy,Orange
3   Joe,Good
4   Pete,High,Anne,Green

I would like to:

• remove commas (replace them by one space)
• wherever I have 2 persons in one cell, insert the "&"symbol after the first person family name and before the second person name.

Desired output:

name
0   John Smith
1   Peter Blue
2   Larry One & Stacy Orange
3   Joe Good
4   Pete High & Anne Green

Tried this code below, but it simply removes commas. I could not find how to insert the "&"symbol in the same code.

df['name']= df['name'].str.replace(r',', '', regex=True) 

Disclaimer: all names in this table are fictitious. No identification with actual persons (living or deceased)is intended or should be inferred.

Answer 1

I would do it following way

import pandas as pd
df = pd.DataFrame({'name':{0: 'John,Smith', 1: 'Peter,Blue', 2:'Larry,One,Stacy,Orange' , 3:'Joe,Good' , 4:'Pete,High,Anne,Green'}})
df['name'] = df['name'].str.replace(',',' ').str.replace(r'(\w+ \w+) ', r'\1 & ', regex=True)
print(df)

gives output

                       name
0                John Smith
1                Peter Blue
2  Larry One & Stacy Orange
3                  Joe Good
4    Pete High & Anne Green

Explanation: replace , s using spaces, then use replace again to change one-or-more word characters followed by space followed by one-or-more word character followed by space using content of capturing group (which includes everything but last space) followed by space followed by & character followed by space.

Answer 2

With single regex replacement:

df['name'].str.replace(r',([^,]+)(,)?', lambda m:f" {m.group(1)}{' & ' if m.group(2) else ''}")

---

0                  John Smith
1                  Peter Blue
2    Larry One & Stacy Orange
3                    Joe Good
4      Pete High & Anne Green

Answer 3

This should work:

import re

def separate_names(original_str):
    spaces = re.sub(r',([^,]*(?:,|$))', r' \1', original_str)
    return spaces.replace(',', ' & ')

df['spaced'] = df.name.map(separate_names)
df

I created a function called separate_names which replaces the odd number of commas with spaces using regex. The remaining commas (even) are then replaced by & using the replace function. Finally I used the map function to apply separate_names to each row. The output is as follows:

Answer 4

In replace statement you should replace comma with space. Please put space between '' -> so you have ' '

df['name']= df['name'].str.replace(r',', ' ', regex=True)
                           inserted space ^ here