CODEWITHSUNDEEP
pythonregex
Drxxd
Drxxd

Reputation: 1929

Finding all of the patterns with RE

I have this string:

"(a) first, (b) second, (c) important"

I'm trying to find all of the strings before the "(c) important" string, so this is my regex:

"(?:\([a-z]\) ([a-z]+), )+\([a-z]\) important"

re.findall finds only the "second" string (without the "first" string).
I tried using re.finditer and the regex module (with the overlapping flag) but they all return the same result.

What should be done here so it will find all of the strings before the "important" string?

Note -
The input string can be different. For example:

"(a) aa, (b) cc, (c) dd, (d) oi, (e) important"  # should return ["aa", "cc", "dd", "oi"]   
"(a) aa, (b) asdf, (c) wer"                      # should return nothing

Upvotes: 4

Views: 47

Answers (1)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626738

You may use

\([a-z]\)\s+([a-z]+)(?=(?:\s*,\s*\([a-z]\)\s+[a-z]+)*\s*,\s*\([a-z]\)\s+important)

See the regex demo

Details

  • \([a-z]\) - a lowercase letter inside parentheses
  • \s+ - 1+ whitespaces
  • ([a-z]+) - Group 1: one or more lowercase letter inside parentheses
  • (?=(?:\s*,\s*\([a-z]\)\s+[a-z]+)*\s*,\s*\([a-z]\)\s+important) - a positive lookahead that matches a location immediately followed by
    • (?:\s*,\s*\([a-z]\)\s+[a-z]+)* - 0 or more repetitions of
      • \s*,\s* - a comma enclosed with 0+ whitespaces
      • \([a-z]\) - a letter enclosed in parentheses
      • \s+ - 1+ whitespaces
      • [a-z]+ - 1+ lowercase letters
    • \s*,\s* - a comma enclosed with 0+ whitespaces
    • \([a-z]\) - a lowercase letter inside parentheses
    • \s+ - 1+ whitespaces
    • important - a word.

Python demo:

import re
strs = ["(a) first, (b) second, (c) important", "(a) aa, (b) cc, (c) dd, (d) oi, (e) important", "(a) aa, (b) asdf, (c) wer" ]
r = re.compile(r'\([a-z]\)\s+([a-z]+)(?=(?:\s*,\s*\([a-z]\)\s+[a-z]+)*\s*,\s*\([a-z]\)\s+important)')
for s in strs:
    print(r.findall(s))

Output:

['first', 'second']
['aa', 'cc', 'dd', 'oi']
[]

Upvotes: 3

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