replace text string in entire column after first occurance-CodePudding

I'm trying to replace all but the first occurrence of a text string in an entire column. My specific case is replacing underscores with periods in data that looks like client_19_Aug_21_22_2022 and I need this to be client_19.Aug.21.22.2022

if I use [1], I get this error: string index out of range
but [:1] does all occurrences (it doesn't skip the first one)
[1:] inserts . after every character but doesn't find _ and replace 

df1['Client'] = df1['Client'].str.replace('_'[:1],'.')

CodePudding user response：

Not the simplest, but solution:

import re
df.str.apply(lambda s: re.sub(r'^(.*?)\.', r'\1_', s.replace('_', '.')))

Here in the lambda function we firstly replace all _ with .. Then we replace the first occurrence of . back with _. And finally, we apply lambda to each value in a column.

CodePudding user response：

Pandas Series have a .map method that you can use to apply an arbitrary function to every row in the Series.

In your case you can write your own replace_underscores_except_first function, looking something like:

def replace_underscores_except_first(s):
    newstring = ''
    # Some logic here to handle replacing all but first.
    # You probably want a for loop with some conditional checking
    return newstring

and then pass that to .map like:

df1['Client'] = df1['Client'].map(replace_underscores_except_first)

CodePudding user response：

An example using map, and in the function check if the string contain an underscore. If it does, split on it, and join back all parts except the first with a dot.

import pandas as pd

items = [
    "client_19_Aug_21_22_2022",
    "client123"
]


def replace_underscore_with_dot_except_first(s):
    if "_" in s:
        parts = s.split("_")
        return f"{parts[0]}_{'.'.join(parts[1:])}"
    return s


df1 = pd.DataFrame(items, columns=["Client"])

df1['Client'] = df1['Client'].map(replace_underscore_with_dot_except_first)
print(df1)

Output

                     Client
0  client_19.Aug.21.22.2022
1                 client123