Groups

Parentheses do two jobs: group things together, and capture what they match so you can use it.

Grouping

>>> re.findall(r"ab+", "ab abb")       # + applies to b only
['ab', 'abb']
>>> re.findall(r"(ab)+", "ab abab")    # + applies to the GROUP
['ab', 'ab']

Hmm — that second result looks odd (['ab', 'ab'] instead of ['ab', 'abab']). That's findall's quirk with groups, explained below. The grouping itself works: (ab)+ matches abab as one match.

Capturing: keep what you matched

Groups numbered from 1, left to right:

>>> m = re.search(r"(\d{4})-(\d{2})-(\d{2})", "date: 2026-10-13")
>>> m.group(1), m.group(2), m.group(3)
('2026', '10', '13')
>>> m.group(0)
'2026-10-13'    # group 0 is the whole match

All verified. This is extraction: pull the year, month, day out of the date separately.

The findall quirk

findall with groups returns the groups, not the whole matches:

>>> re.findall(r"(\d{4})-(\d{2})-(\d{2})", "2026-10-13 and 2025-01-01")
[('2026', '10', '13'), ('2025', '01', '01')]    # tuples of groups!

Verified. If you wanted whole dates, that's the bug — and the fix is non-capturing groups: (?:...) groups without capturing:

>>> re.findall(r"(?:\d{4})-(?:\d{2})-(?:\d{2})", "2026-10-13 and 2025-01-01")
['2026-10-13', '2025-01-01']

Verified. Rule: use (?:...) unless you actually want the captured text. It also reads better — "I'm grouping, not keeping."

Named groups: capture with labels

>>> m = re.search(r"(?P<year>\d{4})-(?P<month>\d{2})", "2026-10")
>>> m.group("year"), m.group("month")
('2026', '10')

Verified. (?P<name>...) — clunky syntax, wonderful readability. In a pattern with five groups, names beat counting parentheses.

Backreferences: match the same text again

\1 means "whatever group 1 matched":

>>> re.search(r"(\w+), \1", "hello, hello").group()
'hello, hello'      # \1 must equal "hello" again
>>> re.search(r"(\w+), \1", "hello, world") is None
True

Verified. Real uses: finding doubled words (\b(\w+) \1\b), matching quote pairs ((['"])(.*?)\1 — the closing quote must match the opening one).

sub: replace using groups

>>> re.sub(r"(\d{4})-(\d{2})-(\d{2})", r"\3/\2/\1", "2026-10-13")
'13/10/2026'

Verified. re.sub replaces matches; \1 in the replacement inserts group 1. Date reformatting in one line — this is the "find-and-replace with superpowers" that sells regex to skeptics.

Next: Alternation