Suppose I have a string that has the same sub-string repeated multiple times and I want to replace each occurrence with a different element from a list.
For example, consider this scenario:
pattern = "_____" # repeated pattern
s = "a(_____), b(_____), c(_____)"
r = [0,1,2] # elements to insert
The goal is to obtain a string of the form:
s = "a(_001_), b(_002_), c(_003_)"
The number of occurrences is known, and the list r has the same length as the number of occurrences (3 in the previous example) and contains increasing integers starting from 0.
I've came up with this solution:
import re
pattern = "_____"
s = "a(_____), b(_____), c(_____)"
l = [m.start() for m in re.finditer(pattern, s)]
i = 0
for el in l:
s = s[:el] + f"_{str(i).zfill(5 - 2)}_" + s[el + 5:]
i += 1
print(s)
Output: a(_000_), b(_001_), c(_002_)
This solves my problem, but it seems to me a bit cumbersome, especially the for-loop. Is there a better way, maybe more "pythonic" (intended as concise, possibly elegant, whatever it means) to solve the task?
You can simply use re.sub() method to replace each occurrence of the pattern with a different element from the list.
import re
pattern = re.compile("_____")
s = "a(_____), b(_____), c(_____)"
r = [0,1,2]
for val in r:
s = re.sub(pattern, f"_{val:03d}_", s, count=1)
print(s)
You can also choose to go with this approach without re using the values in the r list with their indexes respectively:
r = [0,1,2]
s = ", ".join(f"{'abc'[i]}(_{val:03d}_)" for i, val in enumerate(r))
print(s)
a(_000_), b(_001_), c(_002_)