get all occurrences from a matched pattern and split

Question

I'm using the following code to find all the indexes from index 0 to ~ so I could create a row off of each match

import re
s = 'product 1 & product 2|category 1|8~product 4|category 3 |10~product 1 & product 19|category 8|6~product 50|category 4|6'

substring = "~"

matches = re.finditer(substring, s)

matches_positions = [match.start() for match in matches]

print(matches_positions)

output 
[34, 59, 95]

I'm manually using every index in order to show the following output, I would like to create a function that could split and return each occurrence

print(s[0:34])
print(s[34 + 1: 59])
print(s[59 +1 : 95])
print(s[95 +1 : len(s)])


output

product 1 & product 2|category 1|8
product 4|category 3 |10
product 1 & product 19|category 8|6
product 50|category 4|6

Thank you beforehand

Answer 1

You just need to use the str.split built-in function:

outputs = s.split('~')

Here is outputs value:

['product 1 & product 2|category 1|8',
 'product 4|category 3 |10',
 'product 1 & product 19|category 8|6',
 'product 50|category 4|6']