For example, given regex

r"(?P<D>.8)(?P<foo>\d+)(?P<bar>[a-z]+)"

I want to return the dictionary

{'D': r'.8', 'foo': r'\d+', 'bar': r'[a-z]+'}

that maps group name to group regex. Getting the groups is straightforward (with groupindex), but there doesn't seem to be a direct way to get the corresponding regexes.

For example, given regex

r"(?P<D>.8)(?P<foo>\d+)(?P<bar>[a-z]+)"

I want to return the dictionary

{'D': r'.8', 'foo': r'\d+', 'bar': r'[a-z]+'}

that maps group name to group regex. Getting the groups is straightforward (with groupindex), but there doesn't seem to be a direct way to get the corresponding regexes.

• pandas
• regex
• regex-group

2 Answers 2

I believe you mean this am I right

import re

pattern = r"(?P<D>.8)(?P<foo>\d+)(?P<bar>[a-z]+)"

# Regex pattern to capture named groups
group_pattern = repile(r"\(\?P<(?P<name>\w+)>(?P<regex>.*?)\)")

# Extract named groups and their regex patterns
group_patterns = {match.group("name"): match.group("regex") for match in group_pattern.finditer(pattern)}

print(group_patterns)

Check if the regular expression \(\?P<(.+?)>((?:[^)]|(?<=\\)\))+)\) meets your requirements.
An explanation of the regular expression can be found at https://regex101/r/d7MvHp/1

import re

regex = r"\(\?P<(.+?)>((?:[^)]|(?<=\\)\))+)\)"
test_str = ("(?P<D>.8)(?P<foo>\(\d+\))(?P<bar>[a-z]+)")
d = {m.group(1) : m.group(2) for m in re.finditer(regex, test_str)}
print(d)

Output {'D': '.8', 'foo': '\\(\\d+\\)', 'bar': '[a-z]+'}

The snippet can be checked at https://rextester/KTMZ92055