Simplifying a complex regex

Question

I am programmatically generating a very lengthy regex expression in a lua script I'm developing. It works, but at 960 characters, I can't help thinking it can be simplified a bit. However, I can't think of a way to do that without breaking it. In a nutshell, it looks for a string of 1 or more space-separated predefined keywords. A much-shortened version of the regex looks like this. You can see the major portion of it, the keyword list, is repeated.

\(big\|hairy\|regex\)\( \+\(big\|hairy\|regex\)\)*

To remove the duplication, I could use this one. The problem here is that the string starts with a space; whereas, I need it to start with any of the keywords.

\( \+\(big\|hairy\|regex\)\)\+

Of course the next regex doesn't work because \1 refers to the text that was matched and not the regex that found it.

\(big\|hairy\|regex\)\( \+\1\)*

Can the regex be written so that the keyword list appears only once, but still matches correctly?

Answer 1

I would do:

 *\zs\( *\<\(big\|hairy\|regex\)\>\)\+

That I would rewrite:

\v *\zs( *<(big|hairy|regex)>)+

The first * eat the leading spaces if any.

They will not be part of the matching string because of the \zs that define the start of the matching.

The repetition don't force a set of between the keywords but because of the < and > you are forced to have at least a non word but because only are authorized in the regex you should have the desired result.