Regex function to delete all numbers and punctuation except a specified expression?

Question

I have a string in Python:

string = "Hello I am a 21 !string. In section 3.2.F.1.2 we covered 1topic X. On the oth1er hand, in section 1.2.F.1.1 we covered Y. Lastly, in section F.3.2 we 23 covered Z."

I need to remove the random numbers and punctuation from the text such that:

"a 21 !string" --------> "...a string..." and...

"covered 1 topic x." ---------> "covered topic"

My final string should then be:

filtered = "hello i am a string in section 3.2.F.1.2 we covered topic x on the other hand in section 1.2.F.1.1 we covered y lastly in section 1.1.F.3.2 we covered z"

...such that the codes "3.2.F.1.2", "1.2.F.1.1", and "1.1.F.3.2" were not affected by this.

I was able to generate a regex expression to specify the codes with:

regex_codes = "[\d\.]{1,4}F[\.\d]{1,4}"

all_nums_punct = "[0-9 _.,!"'/$]*"

What I cannot figure out is how to "select and remove all numbers and punctuation (all_nums_punct) except these codes (regex_code) pattern".

I tried using a "negative lookahead" pattern to ignore everything that starts with my codes from a previous stackOverflow article, but my selection is not selecting anything.

Answer 1

Using the regex package from the PyPI repository:

import regex

string = "Hello I am a 21 !string. In section 3.2.F.1.2 we covered 1topic X. On the oth1er hand, in section 1.2.F.1.1 we covered Y. Lastly, in section 1.1.F.3.2 we 23 covered Z."
string = regex.sub(r'''[\d\.]{1,4}F[\.\d]{1,4}(*SKIP)(*FAIL)|[0-9_.,!"'/$]''', '', string)
print(string)

Prints:

Hello I am a  string In section F we covered topic X On the other hand in section F we covered Y Lastly in section F we  covered Z

We match either your regex_codes expression or one of your all_nums_punct characters (without the space character). If we match the regex_codes expression, we skip over those characters and fail the test and try the second alternative.

The results will have possibly multiple contiguous space characters. You will need a second replace operation to replace these with a single space:

import regex

string = "Hello I am a 21 !string. In section 3.2.F.1.2 we covered 1topic X. On the oth1er hand, in section 1.2.F.1.1 we covered Y. Lastly, in section 1.1.F.3.2 we 23 covered Z."
string = regex.sub(r'''[\d\.]{1,4}F[\.\d]{1,4}(*SKIP)(*FAIL)|[0-9_.,!"'/$]''', '', string)
string = regex.sub(r' +', ' ', string)
print(string)

Prints:

Hello I am a string In section 3.2.F.1.2 we covered topic X On the other hand in section 1.2.F.1.1 we covered Y Lastly in section 1.1.F.3.2 we covered Z

Update

I will try to answer the question you posed to @WiktorStribiżew concerning how his solution below worked:

re.sub(r"""([.\d]{1,4}F[.\d]{1,4})|[0-9_.,!"'/$]'""", '\1', $string)

Whatever the regular expression matches will be replaced by '\1', which specifies the value of capture group 1. If the regular expression matches a regex_codes, then capture group 1 will be set to what ever it matches and the matched string will be replaced with itself and nothing is modified. However, if the regular expression matches one of the characters you wish to delete, then capture group 1 will be empty and the matched string will be replaced by an empty string. This method does not require the regex package. This method, too, will leave contiguous spaces, which you will probably want to remove as I have indicated.