Regex (Python) - How to match part of a string multiple times?

I need a python regex matching the part of a string multiple times:

My String: aaaa-bb-ccc-dd
My Pattern: ([A-z]+)\-([A-z]+)

I would like to have groups like this:
1: aaaa-bb
2: bb-ccc
3: ccc-dd

Does somebody have an idea on how to do this? If it does not work with regex only, a solution with a python for loop is also very welcome.

Solution

You can use lookahead to get overlapping matches:

(?=\b([A-Za-z]+-[A-Za-z]+)\b)

See the regex demo.

Details:

(?= - start of a positive lookahead that matches a location that is immediately followed with
- \b - a word boundary
- ([A-Za-z]+-[A-Za-z]+) - Group 1: one or more ASCII letters, -, one or more ASCII letters
- \b - a word boundary
) - end of the lookahead.

In Python, use it with re.findall:

import re
text = "aaaa-bb-ccc-dd"
print( re.findall(r'(?=\b([A-Z]+-[A-Z]+)\b)', text, re.I) )
# => ['aaaa-bb', 'bb-ccc', 'ccc-dd']

See the Python demo. Note I changed [A-Za-z] to [A-Z] in the code since I made the regex matching case insensitive with the help of the re.I option. Make sure you are using the r string literal prefix or \b will be treated as a BACKSPACE char, \x08, and not a word boundary.

Variations

(?=\b([^\W\d_]+-[^\W\d_]+)\b) - matching any Unicode letters
(?=(?<![^\W\d_])([^\W\d_]+-[^\W\d_]+)(?![^\W\d_])) - matching any Unicode letters and the boundaries are any non-letters