Regular expressions

Use `re.search` to check if a string contains a sequence of 3 or more consecutive digits. Return True or False.

import re

def has_three_digits(s):
    # your code here
    pass

print(has_three_digits('abc123def'))  # True
print(has_three_digits('abc12def'))   # False
print(has_three_digits('1234'))       # True

import re

def has_three_digits(s):
    return bool(re.search(r'\d{3,}', s))

Use `re.match` to validate that a string is a valid Python identifier (letters, digits, underscores; must start with a letter or underscore).

import re

def is_identifier(s):
    # your code here
    pass

print(is_identifier('my_var'))    # True
print(is_identifier('_private'))  # True
print(is_identifier('2bad'))      # False
print(is_identifier('hello world')) # False

import re

def is_identifier(s):
    return bool(re.match(r'^[A-Za-z_][A-Za-z0-9_]*$', s))

Extract all email addresses from a multi-line string. Return them as a list.

import re

def find_emails(text):
    # your code here
    pass

text = '''
Contact us at [email protected] or [email protected].
You can also reach [email protected] for questions.
'''
print(find_emails(text))
# ['[email protected]', '[email protected]', '[email protected]']

import re

def find_emails(text):
    return re.findall(r'[\w.+-]+@[\w-]+\.[\w.]+', text)

Replace all occurrences of multiple spaces with a single space in a string.

import re

def normalize(s):
    # your code here
    pass

print(normalize('hello   world'))   # 'hello world'
print(normalize('  a  b  c  '))     # ' a b c '

import re

def normalize(s):
    return re.sub(r' {2,}', ' ', s)

Use capturing groups to extract the year, month, and day from a date string in 'DD/MM/YYYY' format.

import re

def parse_date(s):
    # Return (day, month, year) as integers, or None if format is wrong
    pass

print(parse_date('15/03/2024'))  # (15, 3, 2024)
print(parse_date('01/12/1999'))  # (1, 12, 1999)
print(parse_date('invalid'))     # None

import re

def parse_date(s):
    m = re.match(r'^(\d{2})/(\d{2})/(\d{4})$', s)
    if not m:
        return None
    return int(m.group(1)), int(m.group(2)), int(m.group(3))

Use `re.sub` with a function to double all numbers found in a string.

import re

def double_numbers(s):
    # your code here
    pass

print(double_numbers('I have 3 cats and 12 dogs'))  # 'I have 6 cats and 24 dogs'
print(double_numbers('price: 9.99'))                 # 'price: 19.98'

import re

def double_numbers(s):
    return re.sub(r'\d+\.?\d*', lambda m: str(float(m.group()) * 2), s)

Compile a regex pattern once and use it to filter a list of strings, keeping only those that match a phone number pattern: `+X-XXX-XXX-XXXX` (where X is a digit).

import re

def filter_phones(strings):
    # your code here
    pass

items = ['+1-800-555-1234', 'not a phone', '+44-020-555-9876', '123-456']
print(filter_phones(items))  # ['+1-800-555-1234', '+44-020-555-9876']

import re

def filter_phones(strings):
    pattern = re.compile(r'^\+\d+-\d{3}-\d{3}-\d{4}$')
    return [s for s in strings if pattern.match(s)]

Use named groups to parse a log line of the format `[LEVEL] YYYY-MM-DD: message`. Return a dict with 'level', 'date', and 'message' keys.

import re

def parse_log(line):
    # your code here
    pass

print(parse_log('[ERROR] 2024-03-15: Connection refused'))
# {'level': 'ERROR', 'date': '2024-03-15', 'message': 'Connection refused'}
print(parse_log('[INFO] 2024-01-01: Server started'))
# {'level': 'INFO', 'date': '2024-01-01', 'message': 'Server started'}

import re

def parse_log(line):
    m = re.match(r'^\[(?P<level>\w+)\] (?P<date>\d{4}-\d{2}-\d{2}): (?P<message>.+)$', line)
    return m.groupdict() if m else None

Use `re.split` to split a string on any punctuation mark (`.`, `,`, `!`, `?`, `;`, `:`). Filter out empty strings from the result.

import re

def split_on_punctuation(s):
    # your code here
    pass

print(split_on_punctuation('Hello, world! How are you?'))
# ['Hello', ' world', ' How are you']
print(split_on_punctuation('one.two;three:four'))
# ['one', 'two', 'three', 'four']

import re

def split_on_punctuation(s):
    return [part for part in re.split(r'[.,!?;:]', s) if part]

Use `re.VERBOSE` (or `re.X`) flag to write a readable, commented regex that validates a strong password: at least 8 chars, at least one uppercase, one lowercase, one digit, one special char (`!@#$%^&*`).

import re

def is_strong_password(s):
    # your code here
    pass

print(is_strong_password('Abc@1234'))    # True
print(is_strong_password('abcdefgh'))    # False (no uppercase/digit/special)
print(is_strong_password('Abc@12'))      # False (too short)

import re

def is_strong_password(s):
    pattern = re.compile(r'''
        ^                    # start of string
        (?=.*[A-Z])          # at least one uppercase
        (?=.*[a-z])          # at least one lowercase
        (?=.*\d)             # at least one digit
        (?=.*[!@#$%^&*])    # at least one special char
        .{8,}               # at least 8 characters total
        $                   # end of string
    ''', re.VERBOSE)
    return bool(pattern.match(s))