Python, RegEx, Replace a certain part of a match

Question

I am trying to replace a certain part of a match that a regex found. The relevant strings have the following format:

"<Random text>[Text1;Text2;....;TextN]<Random text>"

So basically there can be N Texts seperated by a ";" inside the brackets. My goal is to change the ";" into a "," (but only for the strings which are in this format) so that I can keep the ";" as a seperator for a CSV file. So the result should be:

"<Random text>[Text1,Text2,...,TextN]<Random text>"

I can match the relevant strings with something like

re.compile(r'\[".*?((;).*?){1,4}"\]')

but if I try to use the sub method it replaces the whole string.

I have searched stackoverflow and I am pretty sure that "capture groups" might be the solution but I am not really getting there. Can anyone help me?

I ONLY want to change the ";" in the ["Text1;...;TextN"]-parts of my text file.

Text1, Text2, ... Text N are placeholders. So the string can e.g. look like ["Hello;Bye;This123"]. I know where its starts due to [" and where it ends due to "]. — python1235432
– python1235432, Commented Jan 21, 2020 at 9:32

Gurmanjot Singh · Accepted Answer · 2020-01-21 09:40:48Z

6

Try this regex:

;(?=(?:(?!\[).)*])

Replace each match with a ,

Click for Demo

Explanation:

; - matches a ;
(?=(?:(?!\[).)*]) - makes sure that the above ; is followed by a closing ] somewhere later in the string but before any opening bracket [
- (?=....) - positive lookahead
- (?:(?!\[).)* - 0+ occurrences of any character which does not start with [
- ] - matches a ]

edited Jan 21, 2020 at 9:40

answered Jan 21, 2020 at 9:34

Gurmanjot Singh

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Comments

The fourth bird · Accepted Answer · 2020-01-21 10:41:06Z

If you want to match a ; before a closing ] and not matching [ in between you could use:

;(?=[^[]*])

; Match literally
(?= Positive lookahead, assert what is on the right is
- [^[]* Negated character class, match 0+ times any char except [
] Match literally
) Close lookahead

Regex demo

Note that this will also match if there is no leading [

If you also want to make sure that there is a leading [ you could make use of the PyPi regex module and use \G and \K to match a single ;

(?:\[(?=[^[\]]*])|\G(?!^))[^;[\]]*\K;

Regex demo | Python demo

import regex

pattern = r"(?:\[(?=[^[\]]*])|\G(?!^))[^;[\]]*\K;"
test_str = ("[\"Text1;Text2;....;TextN\"];asjkdjksd;ajksdjksad[\"Text1;Text2;....;TextN\"]\n\n"
    ".[\"Text1;Text2\"]...long text...[\"Text1;Text2;Text3\"]....long text...[\"Text1;...;TextN\"]...long text...\n\n"
    "I ONLY want to change the \";\" in the [\"Text1;...;TextN\"]")

result = regex.sub(pattern, ",", test_str)
print (result)

Output

["Text1,Text2,....,TextN"];asjkdjksd;ajksdjksad["Text1,Text2,....,TextN"]

.["Text1,Text2"]...long text...["Text1,Text2,Text3"]....long text...["Text1,...,TextN"]...long text...

I ONLY want to change the ";" in the ["Text1,...,TextN"]

KAMLESH KUMAR · Accepted Answer · 2020-01-21 10:15:31Z

1

You can try this code sample:

import re
x = 'anbhb["Text1;Text2;...;TextN"]nbgbyhuyg["Text1;Text2;...;TextN"][]nhj,kji,'
for i in range(len(x)):
    if x[i] == '[' and x[i + 1] == '"':
        while x[i+2] != '"':
            list1 = list(x)
            if x[i] == ';':
                list1[i] = ','
                x = ''.join(list1)

            i = i + 1

print(x)

edited Jan 21, 2020 at 10:15

answered Jan 21, 2020 at 9:25

KAMLESH KUMAR

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1 Comment

python1235432 Over a year ago

If I use str.replace(";",",") it will change every ";" in my text-file. I only want to replace the ";" in the parts which have the following format: ["Text1;Text2;Text3;..;TextN"]

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