CODEWITHSUNDEEP
regex
BenH
BenH

Reputation: 35

How to exclude a pattern that starts with a phrase, but the pattern may contain substring of the phrase?

With this example,

whether or not apple
apple
not apple

How could I find either apple or not apple and exclude whether or not apple? My attempt was to use negative lookbehind like

(?!whether or not)(([Nn]ot)?\s*apple)

but this finds all as

Upvotes: 2

Views: 825

Answers (2)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 627536

As your consuming pattern starts with an optional pattern, you need to account for these two scenarios and implement two separate lookbehinds:

(?<!\bwhether or )(?<!\bwhether or not )(?:\bnot )?\bapple\b

See the regex demo. Details:

  • (?<!\bwhether or ) - no whether or allowed immediately on the left
  • (?<!\bwhether or not ) - no whether or not allowed immediately on the left
  • (?:\bnot )? - an optional sequence of not word and a space
  • \bapple\b - a whole word apple.

Another approach is to match the string you want to skip and skip it:

\bwhether\s+or\s+not\s+apple\b(*SKIP)(*F)|(?:\bnot\s+)?\bapple\b

See this regex demo. Here, \bwhether\s+or\s+not\s+apple\b matches the whether or not apple and (*SKIP)(*F) drop the match, skips to the position of the failure and restarts a new match search.

Upvotes: 1

Artyom Vancyan
Artyom Vancyan

Reputation: 5388

The pattern you are looking for is ^(?:not\s)?apple

Python

import re

re.match(r"^(?:not\s)?apple", "apple")  # matches
re.match(r"^(?:not\s)?apple", "not apple")  # matches
re.match(r"^(?:not\s)?apple", "whether or not apple")  # does not match

JavaScript

/^(?:not\s)?apple/.test("apple"); // matches
/^(?:not\s)?apple/.test("not apple"); // matches
/^(?:not\s)?apple/.test("whether or not apple"); // does not match

Upvotes: 0

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