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String patterns %b() exclude \(\)

 
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sir-gunny
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PostPosted: Wed Feb 21, 2024 3:58 am    Post subject: String patterns %b() exclude \(\) Reply with quote
Hi.

Does anyone have any idea how I can exclude \(text\) from string:gsub("%b()","")?

Code:
s = [[\(keep (this not) this\) Hello (remove ( me)please) world]]
print(s:gsub("%b()",""))


Output:
Code:
\ Hello  world


But i need this:
Code:
\(keep this\) Hello world


Thx
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AylinCE
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PostPosted: Wed Feb 21, 2024 9:23 am    Post subject: Reply with quote
Code:
s = [[\(keep (this not) this\) Hello (remove ( me)please) world \(keep (this not) this\)]]
res, repcount = string.gsub(s, "( %b())", "")
print(res) --> \(keep this\) Hello world
print(repcount)


--res--> \(keep this\) Hello world \(keep this\)
--repcount--> 3
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ParkourPenguin
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PostPosted: Wed Feb 21, 2024 2:04 pm    Post subject: This post has 1 review(s) Reply with quote
More generally, use a frontier pattern to exclude parenthesis preceded by a backslash:
Code:
local s = [[\(keep (this not) this\) Hello (remove ( me)please) world]]
print((s:gsub('%f[\\(]%b()','')))

-- \(keep  this\) Hello  world
Note that there are 2 spaces where the substitution occurred. Deal with that as you see fit.
If you always want there to be at least one space, perhaps this is fine:
Code:
local s = [[\(keep (this not) this\) Hello (remove ( me)please) world]]
print((s:gsub(' ?%f[\\(]%b() ?',' ')))

-- \(keep this\) Hello world

If you need more advanced logic, use a function to figure it out.
Code:
local s = [[\(keep (this not) this\) Hello (remove ( me)please) world]]
print((s:gsub('(.?)%f[\\(]%b()(.?)',function(pre, post)
  if pre == ' ' and post == ' ' then
    return ' '
  else
    return ''
  end
end)))

-- \(keep this\) Hello world

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AylinCE
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PostPosted: Wed Feb 21, 2024 3:45 pm    Post subject: Reply with quote
I thought I provided a solution that produced the desired result.

But the lesson to be learned from this is:

Exclude:
Code:
%f[] , %f()

Delete:
Code:
%b(), %b[], %b<> -- etc.


Oki ..
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sir-gunny
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PostPosted: Thu Feb 22, 2024 4:14 am    Post subject: Reply with quote
Hi.

Thank you very much for your help. Unfortunately my question was not clear enough, sorry. The "()" part can be anywhere in the string e.g. at the beginning, between words, after special characters etc. . Therefore ParkourPenguin's solution is ideal for me. Until now I did not know frontier pattern. Thanks for that! Smile
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panraven
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PostPosted: Thu Feb 22, 2024 3:53 pm    Post subject: Reply with quote
If you can find an unique literal string (z) NOT in the string to be search, eg chr(0), then you can gsub '\(' with z, do your need gsub, and finally gsub z with the original. You may find the unique z by other way if cannot be sure a fixed string.

UPDATED:
Actually, above is not working.
1. the str like '\(' need to be escape for pattern matching, ie ->'\%(';
2. chr(0) seems special, cause garbage result, not know why, but using chr(1), chr(2) etc ok;
Code:

local z1, z2, r1, r2 = string.char(1), string.char(2), [[\%(]], [[\%)]]
print(
s:gsub(r1,z1)
 :gsub(r2,z2)
 :gsub('%b()','')
 :gsub(z2,r2)
 :gsub(z1,r1)
)


Also, it is not a general solution, only work for removing %b();
if it need to process content in %b(), then the result may not correct as %b() may contain replaced char.
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sir-gunny
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PostPosted: Fri Feb 23, 2024 3:40 am    Post subject: Reply with quote
My first solution was similar. I just replaced the characters with "-#" and "#-". But I suspected that there was a more "elegant" solution.

In my case, the content between "()" only consists of information from the programmers. This is not displayed in the game.
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