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

• whether or not apple
• apple
• not apple

CodePudding user response：

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.

CodePudding user response：

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