The Regular Expression "r\d{1,2}" will match the character r and any digit after that (but for a count in between 1 and 2 (\d{1,2}) and replace them by p0

What if there is "r" with more than 2 numbers after it (e.i. r463718/r3/4/r5)?
Is there a way to get the same result no matter the amount of numbers after the "r"

Is there something more elegant than ("r\d{1,9999}","p0" ) ?
Or would that even work the way I think it would.

You could do something a bit simpler in this case without regex!

You could likely just do Repeat(â€śp0/â€ś, column.split(â€ś/â€ś).count())

All that is doing is counting how many â€śr/â€ś things you have regardless of the number of digits after the r and then repeating your p0 string a matching number of times