A way to go is to match the parts around the comma and then remove it:

var input = '124,000 km';
input.replace(/(\d+),(\d+\s?km)/g, '2');

 public static void main(String args[]) throws IOException

    {

        String regex = "(?<=[\\d])(,)(?=[\\d])";
        Pattern p = Pattern.compile(regex);
        String str = "John loves cakes and he always orders them by dialing \"989,444 1234\". Johns credentials are as follows\" \"Name\":\"John\", \"Jr\", \"Mobile\":\"945,234,1110\"";
        Matcher m = p.matcher(str);
        str = m.replaceAll("");
        System.out.println(str);
    }

Output

John loves cakes and he always orders them by dialing "989444 1234". Johns credentials are as follows" "Name":"John", "Jr", "Mobile":"9452341110"

You could replace anything with the pattern (comma followed by a number) with the number itself.

x <- "I want to see 102,345,5 dogs, but not too soo; it's 3,242 minutes away"
gsub(",([[:digit:]])", "\\1", x)
#[1] "I want to see 1023455 dogs, but not too soo; it's 3242 minutes away"
#or
gsub(",([0-9])", "\\1", x)
#[1] "I want to see 1023455 dogs, but not too soo; it's 3242 minutes away"

You may use any of the solutions below:

var s = "abc,def,2,100,xyz!,:))))";
Console.WriteLine(Regex.Replace(s, @"(\d),(\d)", "2"));   // Does not handle 1,2,3,4 cases
Console.WriteLine(Regex.Replace(s, @"(\d),(?=\d)", "$1"));   // Handles consecutive matches with capturing group+backreference/lookahead
Console.WriteLine(Regex.Replace(s, @"(?<=\d),(?=\d)", ""));  // Handles consecutive matches with lookbehind/lookahead, the most efficient way
Console.WriteLine(Regex.Replace(s, @",(?<=\d,)(?=\d)", "")); // Also handles all cases

See the C# demo.

Explanations:

• (\d),(\d) - matches and captures single digits on both sides of , and 2 are replacement backreferences that insert captured texts back into the result
• (\d),(?=\d) - matches and captures a digit before ,, then a comma is matched and then a positive lookahead (?=\d) requires a digit after ,, but since it is not consumed, onyl $1 is required in the replacement pattern
• (?<=\d),(?=\d) - only such a comma is matched that is enclosed with digits without consuming the digits ((?<=\d) is a positive lookbehind that requires its pattern match immediately to the left of the current location)
• ,(?<=\d,)(?=\d) - matches a comma and only after matching it, the regex engine checks if there is a digit and a comma immediately before the location (that is after the comma), and if the check if true, the next char is checked for a digit. If it is a digit, a match is returned.

RegexHero.net test:

Bonus:

You may just match a pattern like yours with \d,\d and pass the match to the MatchEvaluator method where you may manipulate the match further:

Console.WriteLine(Regex.Replace(s, @"\d,\d", m => m.Value.Replace(",",string.Empty))); // Callback method

Here, m is the match object and m.Value holds the whole match value. With .Replace(",",string.Empty), you remove all commas from the match value.

You can simplify the regular expression:

num.replace(/,/g, '')

• Find what: \$(\d+),(\d+)(?:,(\d+))?
• Replace with: \1\2\3
• Search mode: Regular expression

Explanation:

• \$ matches the characther $
• \d+ matches one or more digits. 1st capturing group
• , matches the comma
• \d+ matches one or more digits. 2nd capturing group
• (?:,(\d+))? Non-capturing group that optionally matches the comma followed by one or more digits (3rd capturing group)

Use a zero-width negative lookahead to make sure the to be replaced substrings (commas here) are not followed by {space(s)}{digit} at the end:

,(?!\s+\d$)

Example:

In [227]: text = '52A, XYZ Street, ABC District, 2'

In [228]: re.sub(',(?!\s+\d$)', '', text)
Out[228]: '52A XYZ Street ABC District, 2'

Edit:

If you have more commas after the ,{space(s)}{digit} substring, and want to keep them all, leverage a negative lookbehind to make sure the commas are not preceded by {space}{digit<or>[A-Z]}:

(?<!\s[\dA-Z]),(?!\s+\d,?)

Example:

In [229]: text = '52A, XYZ Street, ABC District, 2, M, Brown'

In [230]: re.sub('(?<!\s[\dA-Z]),(?!\s+\d,?)', '', text)
Out[230]: '52A XYZ Street ABC District, 2, M, Brown'

In [231]: text = '52A, XYZ Street, ABC District, 2'

In [232]: re.sub('(?<!\s[\dA-Z]),(?!\s+\d,?)', '', text)
Out[232]: '52A XYZ Street ABC District, 2'

Parsing CSV should be done with a proper csv parser. I would recommend perl as well.

perl -MText::ParseWords -ne '
    @line = parse_line(",", 1, $_); 
    print join "," , map { s/,//g if $_ =~ /^[0-9,"]+$/; $_ } @line
' text.csv

Test:

$ cat text.csv
1,2,"12,345",x,y,"a,b"
"a,c","12,345",x,y,"a,b"

$ perl -MText::ParseWords -ne '
    @line = parse_line(",", 1, $_);
    print join "," , map { s/,//g if $_ =~ /^[0-9,"]+$/; $_ } @line
' text.csv
1,2,"12345",x,y,"a,b"
"a,c","12345",x,y,"a,b"

To make in-place changes you can use -i option or re-direct the output to another file.

You may use capturing groups to retain digits:

(\$\d+),(\d+)

and replace with $1$2. You may remove \$ if you do not care if it is a currency or not.

The (\$\d+),(\d+) regex matches:

• (\$\d+) - Group 1 matching $ as a literal symbol followed with 1 or more digits
• , - a literal comma
• (\d+) - Group 2 matching 1 or more digits

The $1 and $2 are backreferences that retrieve the texts stored in the memoru buffers for both groups.

/

Note that there are other ways to do the same, you can use lookarounds or a regex with \K, or using both, but capturing seems to me the most efficient solution for this case.

Your original regex does not match a comma with [\d.] pattern inside the lookbehind. If you add a comma there, it will work.

However, a more generic pattern can turn out more future-proof:

(?<=\\num\{[^{}]*),(?=[^{}]*\})

See the regex demo.

Now, it matches

• (?<=\\num\{[^{}]*) - a positive lookbehind that matches a location that is immediately preceded with \nun{ and then any zero or more chars other than { and }
• , - a comma
• (?=[^{}]*\}) - a positive lookahead that matches a location that is immediately followed with any zero or more chars other than { and } and then a }.