CODEWITHSUNDEEP
pythonregexpython-3.x
MJB
MJB

Reputation: 873

python regex to find alphanumeric string with at least one letter

I am trying to figure out the syntax for regular expression that would match 4 alphanumeric characters, where there is at least one letter. Each should be wrapped by: > and < but I wouldn't like to return the angle brackets.

For example when using re.findall on string >ABCD<>1234<>ABC1<>ABC2 it should return ['ABCD', 'ABC1'].

1234 - doesn't have a letter

ABC2 - is not wrapped with angle brackets

Upvotes: 2

Views: 4107

Answers (2)

M Sandler
M Sandler

Reputation: 11

import re
sentence = ">ABCD<>1234<>ABC1<>ABC2"
pattern = "\>((?=[a-zA-Z])(.){4})\<"
m = [m[0] for m in re.findall(pattern, sentence)]
#outputs ['ABCD', 'ABC1']

Upvotes: 0

anubhava
anubhava

Reputation: 784898

You may use this lookahead based regex in python with findall:

(?i)>((?=\d*[a-z])[a-z\d]{4})<

RegEx Demo

Code:

>>> regex = re.compile(r">((?=\d*[a-z])[a-z\d]{4})<", re.I)
>>> s = ">ABCD<>1234<>ABC1<>ABC2"
>>> print (regex.findall(s))
['ABCD', 'ABC1']

RegEx Details:

  • re.I: Enable ignore case modifier
  • >: Match literal character >
  • (: Start capture group
    • (?=\d*[a-z]): Lookahead to assert we have at least one letter after 0 or more digits
    • [a-z\d]{4}: Match 4 alphanumeric characters
  • ): End capture group
  • <: Match literal character <

Upvotes: 5

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