regex find substring and replace characters and update whole string

Use regex capture groups:

import re
s= 'this is, not good "type of ,question" to ask, on stackoverflow'
re.sub(r'(".*?),(.*?")', r'\1\2', s)

output:

'this is, not good "type of question" to ask, on stackoverflow'

Explanation: (stuff) in regex stands for capture groups, \1 and \2 respectively substitutes the part before and after the , character within the quoted part of string. Please note this also works for multiple quotes within a single string as well.

Another way that gives you some flexibility is you can do it by two steps:

1. Find all the matches that are contained in quotations,

2. In each match look for and replace the ','.

Example:

# define a pattern that gets you everything inside a double quote
pat = re.compile(r'"[^"]+"')

# re.sub the quote pattern and replace the , in each of those matches.
string = pat.sub(lambda x: x.group(0).replace(',',''), string)

# 'this is, not good "type of question" to ask, on stackoverflow'

The flexibility of this is it allows you to replace as many ',' as you need, and you can perform other changes as well once you have located all the double quote patterns.

How about a combination of split() and replace()? :

s = 'this is, not good "type of ,question" to ask, on stackoverflow'

splitted = s.split('"')
print(s.replace(splitted[1], splitted[1].replace(',', '')))

# this is, not good "type of question" to ask, on stackoverflow

Note: This works in this case, but does not work in cases where you have the exact same string within double quotes outside the double quotes.

How about this:

b=""" "hello, howdy". sample text, text then comes "Another, double, quotes" """

for str_match in re.findall(r"\".*?\"",b):
    b = re.sub(str_match,re.sub(r","," ",str_match),b)

print(b)

output: "hello howdy". sample text, text then comes "Another double quotes" '

I'm not completely sure if this will match all your requirements, but on the template you offer the following would return what you are looking for.

result = re.sub('("(?:[^"])*),((?:[^"])*")', r"\1 \2")