Re: how to translate lua pattern "(.*)and(.*)" with lpeg re ?

[albertmcchan <albertmcchan@...>]

>> On Feb 16, 2018, at 3:17 PM, Dirk Laurie <dirk.laurie@gmail.com> wrote:
>> 
>> OK, so the assignment becomes "find the last occurrence of a keyword
>> and return the strings before and after it." I'm not too proud to
>> admit that I would do that with a mixture of Lua and LPeg. Too late at
>> night in my timezone to work out the details.
> 
> -- You can do above in lua easily, with the help of xpattern.lua
> -- I post a copy in https://github.com/achan001/LPeg-anywhere
> 
> xpattern = require 'xpattern'
> X = xpattern.P
> p1 = X'(.*)and%s+' * (X'possibly' + 'likely' + 'definitely') * X'%s+(.*)'
> p1_match = p1:compile()
> 
> t = 'lpeg and possibly xpattern will work, and definitely worth trying'
> = p1_match(t)
> lpeg and possibly xpattern will work,                  worth trying
> 
> -- now, a plug for my patched lpeg-anywhere :-)
> re = require 're'
> z = re.compile[[ 'and' %s+ ('possibly' / 'likely' / 'definitely') %s+ ]]
> p2 = re.compile("{(. >&%z)*} %z {.*}", {z=z})
> = p2:match(t)
> lpeg and possibly xpattern will work,                  worth trying
> 
> -- '>%pat' == '(g <- %pat / .[^%pat]* g)',  and I meant it literally.
> -- [^%pat] = [^a], by actually examine the lpeg object %pat for head-chars
> --> [^%pat]* guaranteed the text begins with 'a' for next match
> --> less false start means better performance.
> 
> on my old pentium 3, 1 million iterations
> p1_match take 8.903 sec
> p2: match take 6.209 sec
> 

fix a few typos ...