وحدة إعادة: regex في بيثون

import re
# raw: \d is a single token
pattern = r"\d+"
# NOT raw: you would have to write "\\d+" otherwise Python interprets \d as "d"

import re
m = re.search(r"\d+", "ho 42 mele e 17 pere")
m.group()    # '42'
m.start()    # 3
m.end()      # 5

if m := re.search(r"\d+", testo):
    print(m.group())

re.match(r"\d+", "42 anni")     # match!
re.match(r"\d+", "ho 42 anni")  # None (doesn't start with a digit)

re.findall(r"\d+", "ho 42 mele, 17 pere, 3 banane")
# ['42', '17', '3']

re.findall(r"(\w+)@(\w+)", "ada@x bob@y")
# [('ada', 'x'), ('bob', 'y')]

re.sub(r"\d+", "###", "ho 42 mele e 17 pere")
# 'ho ### mele e ### pere'

re.sub(r"\d+", lambda m: str(int(m.group()) * 2), "ho 42 mele")
# 'ho 84 mele'

m = re.search(r"(?P<anno>\d{4})-(?P<mese>\d{2})-(?P<giorno>\d{2})", "2025-01-15")
m.group("anno")    # '2025'
m.groupdict()      # {'anno': '2025', 'mese': '01', 'giorno': '15'}

re.search(r"ciao", "CIAO", re.IGNORECASE)       # case-insensitive
re.findall(r"^.+$", testo, re.MULTILINE)        # ^/$ match every line
re.search(r"a.b", "a\nb", re.DOTALL)            # . matches newline too