How to replace specific digits in column using python pandas.

data['value'].str[3:5].replace('10','00') returns a pd.Series consisting of each string sliced by [3:5] with the replace method applied in each row. What you're looking for is returning the whole string with replace applied in each row that matches your condition, which can be achieved like this:

import pandas as pd

data = pd.DataFrame({ # small part of your DF
    'value': ['34 02:0', '36 06:0', '28 10:0'], # Third row should be changed
    'temp1': [16.0, 8.0, 18.0],
    'temp2': [17, 27, 21] 
})

mask = data['value'].str[3:5] == '10'
data.loc[mask, 'value'] = data.loc[mask, 'value'].str.replace('10', '00')

>>> print(data)
     value  temp1  temp2
0  34 02:0   16.0     17
1  36 06:0    8.0     27
2  28 00:0   18.0     21 # Third row changes, yay!

This code could be introducing a bug if there's a value that matches '10' more than once, for example 10 10:0. You can solve this by calling .replace('10:', '00:') instead.

You can also just use regex to match something like r'\s10\:' and call .replace(re.compile(r'\s10\:'), ' 00:').

import re

r = re.compile(r'\s10\:')
data['value'] = data['value'].str.replace(r, ' 00:', regex=True) # no need to define a condition at all

>>> print(data)
     value  temp1  temp2
0  34 02:0   16.0     17
1  36 06:0    8.0     27
2  28 00:0   18.0     21

This last solution is not as explicit as the first one with your condition.