Regular Expressions 101

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An explanation of your regex will be automatically generated as you type.
Detailed match information will be displayed here automatically.
  • All Tokens
  • Common Tokens
  • General Tokens
  • Anchors
  • Meta Sequences
  • Quantifiers
  • Group Constructs
  • Character Classes
  • Flags/Modifiers
  • Substitution
  • A single character of: a, b or c
    [abc]
  • A character except: a, b or c
    [^abc]
  • A character in the range: a-z
    [a-z]
  • A character not in the range: a-z
    [^a-z]
  • A character in the range: a-z or A-Z
    [a-zA-Z]
  • Any single character
    .
  • Alternate - match either a or b
    a|b
  • Any whitespace character
    \s
  • Any non-whitespace character
    \S
  • Any digit
    \d
  • Any non-digit
    \D
  • Any word character
    \w
  • Any non-word character
    \W
  • Non-capturing group
    (?:...)
  • Capturing group
    (...)
  • Zero or one of a
    a?
  • Zero or more of a
    a*
  • One or more of a
    a+
  • Exactly 3 of a
    a{3}
  • 3 or more of a
    a{3,}
  • Between 3 and 6 of a
    a{3,6}
  • Start of string
    ^
  • End of string
    $
  • A word boundary
    \b
  • Non-word boundary
    \B

Regular Expression
Processing...

Test String

Substitution
Processing...

Code Generator

Language

Generated Code

import re regex = re.compile(r"^!*(?P<title>[A-Za-z0-9].+?)(?:\s+(?P<isbn>[0-9]{10}))?\s+by\s+(?P<author>[A-Z0-9][^(]+)(?=\s\()", flags=re.MULTILINE) test_str = ("!!Mixed Fortunes - An Economic History of China Russia and the West 0198703635 by Vladimir Popov (Jun 17, 2014 4_1).pdf\n" "!Mixed Fortunes - An Economic History of China Russia and the West 0198703635 by Vladimir Popov (Jun 17, 2014 4_1).pdf\n" "Mixed Fortunes - An Economic History of China Russia and the West 0198703635 by Vladimir Popov (Jun 17, 2014 4_1).pdf\n" "!!Mixed Fortunes - An Economic History of China Russia and the West by Vladimir Popov (Jun 17, 2014 4_1).pdf\n" "!!Mixed Fortunes - An Economic History of China Russia and the West by 1 Vladimir Popov (Jun 17, 2014 4_1).pdf") subst = "\\3" result = regex.sub(subst, test_str) if result: print(result)

Please keep in mind that these code samples are automatically generated and are not guaranteed to work. If you find any syntax errors, feel free to submit a bug report. For a full regex reference for Python, please visit: https://docs.python.org/3/library/re.html