Capturing repeated pattern in Python

You want to match consectuively repeating \033[\d+m chunks of text and join the numbers after [ with a semi-colon.

You may use

re.sub(r'(?:\\033\[\d+m){2,}', lambda m: r'\033['+";".join(set(re.findall(r"\[(\d+)", m.group())))+'m', text)

See the Python demo online

The (?:\\033\[\d+m){2,} pattern will match two or more sequences of \033[ + one or more digits + m chunks of texts and then, the match will be passed to the lambda expression, where the output will be: 1) \033[, 2) all the numbers after [ extracted with re.findall(r"\[(\d+)", m.group()) and deduplicated with the set, and then 3) m.

The patterns in the string to be modified have not been made clear from the question. For example, is 033 fixed or might it be 025 or even 25? I've made certain assumptions in using the regex

r" ^(\\0(\d+)\[\2)[a-z]\\0\2\[(\d[a-z].+)

to obtain two capture groups that are to be combined, separated by a semi-colon. I've attempted to make clear my assumptions below, in part to help the OP modify this regex to satisfy alternative requirements.

Demo

The regex performs the following operations:

^           # match beginning of line
(           # begin cap grp 1
  \\0       # match '\0'
  (\d+)     # match 1+ digits in cap grp 2
  \[        # match '['
  \2        # match contents of cap grp 2
)           # end cap grp 1
[a-z]       # match a lc letter
\\0         # match '\0'      
\2          # match contents of cap grp 2
\[          # match '['
(\d[a-z].+) # match a digit, then lc letter then 1+ chars to the
            #   end of the line in cap grp 3

As you see, the portion of the string captured in group 1 is

\033[33

I've assumed that the part of this string that is now 033 must be two or more digits beginning with a zero, and the second appearance of a string of digits consists of the same digits after the zero. This is done by capturing the digits following '0' (33) in capture group 2 and then using a back-reference \2.

The next part of the string is to be replaced and therefore is not captured:

m\\033[

I've assumed that m must be one lower case letter (or should it be a literal m?), the backslash and zero and required and the following digits must again match the content of capture group 2.

The remainder of the string,

1mThis is a warning\033[0m: Virus manager \033[4mfailed\033[0m\033[0m

is captured in capture group 3. Here I've assumed it begins with one digit (perhaps it should be \d+) followed by one lower case letter that needn't be the same as the lower case letter matched earlier (though that could be enforced with another capture group). At that point I match the remainder of the line with .+, having given up matching patterns in that part of the string.

One may alternatively have just two capture groups, the capture group that is now #2, becoming #1, and #2 being the part of the string that is to be replaced with a semicolon.

This is pretty straightforward for the cases you desribe here; simply write out from left to right what you want to match and capture. Repeating capturing blocks won't help you here, because only the most recently captured values would be returned as a result.

\\033\[(\d+)m\\033\[(\d+)m