How would I search a table for a close enough match to a keyword for example,
The keyword would be "Volt" and
Table = {"Guest 12930", "Voltoxus", "Player", "Noob"}
How would I search that table and return the word that it most likely resembles please help.
You haven't really defined what "close enough" means. Here's a few possible definitions
searchsearchThe first two approaches have more obvious drawbacks.
Contains
Length of LCS
I think Levenshtein distance is the most promising.
It has a simple recursive definition (taken from Rosetta Code)
function Levenshtein_Distance( s1, s2 )
if s1:len() == 0 then return s2:len() end
if s2:len() == 0 then return s1:len() end
if s1:sub( -1, -1 ) == s2:sub( -1, -1 ) then
return Levenshtein_Distance( s1:sub( 1, -2 ), s2:sub( 1, -2 ) )
end
local a = Levenshtein_Distance( s1:sub( 1, -2 ), s2:sub( 1, -2 ) )
local b = Levenshtein_Distance( s1:sub( 1, -1 ), s2:sub( 1, -2 ) )
local c = Levenshtein_Distance( s1:sub( 1, -2 ), s2:sub( 1, -1 ) )
if a > b then return b + 1 end
if a > c then return c + 1 end
return a + 1
end
Unfortunately this is incredibly inefficient because of memoization reasons.
A "dynamic programming" approach could solve this, or we can apply memoization (save previously computed values).
local LevenshteinMemoize = {};
function Levenshtein_Distance( s1, s2 )
if #s1 < #s2 then
return Levenshtein_Distance(s2, s1)
end
LevenshteinMemoize[s1] = LevenshteinMemoize[s1] or {}
if LevenshteinMemoize[s1][s2] then
return LevenshteinMemoize[s1][s2]
end
if s1:len() == 0 then return s2:len() end
if s2:len() == 0 then return s1:len() end
if s1:sub( -1, -1 ) == s2:sub( -1, -1 ) then
return Levenshtein_Distance( s1:sub( 1, -2 ), s2:sub( 1, -2 ) )
end
local a = Levenshtein_Distance( s1:sub( 1, -2 ), s2:sub( 1, -2 ) )
local b = Levenshtein_Distance( s1:sub( 1, -1 ), s2:sub( 1, -2 ) )
local c = Levenshtein_Distance( s1:sub( 1, -2 ), s2:sub( 1, -1 ) )
if a > b then
R = b + 1
elseif a > c then
R = c + 1
else
R = a + 1
end
LevenshteinMemoize[s1][s2] = R
return R
end
Now we can find the string most similar by the smallest Levenshtein_Distance:
best = 1
keyword = "volt"
keyword = keyword:lower()
for i = 2, #list do
if Levenshtein_Distance(list[i]:lower(), keyword) <
Levenshtein_Distance(list[best]:lower(), keyword) then
best = i
end
end
-- best match is list[best]
Function straight out of the VoteKicker script provided by Roblox.
local players = {"Guest 12930", "Voltoxus", "Player", "Noob"}
function matchPlayer(str)
-- find all players that start with the str
-- if there is only one, match it
-- 0 or 2+, don't match it
local result = nil
for i = 1,#players do
if (string.find(string.lower(players[i]), str) == 1) then
if (result ~= nil) then return nil end
result = players[i]
end
end
return result
end
To call the function, you do
matchPlayer(string)
This is a general example of what you may want to use:
-- Let's say this variable is an inputed value that is being searched for:
local search = "volT" -- Also, case doesn't matter.
local result = matchPlayer(search)
print(result) -- If there is a match, this should return the name.
print(result ~= nil) -- This will print true if it exists, and false if it doesn't.
Returns nil every time
Well you have to set the function at the beginning of the function.
You could try (just for example an admin check)
local admins = {"Guest 12930", "Voltoxus", "Player", "Noob"}
getPermission = function(name)
local admin = false
name = string.lower(name)
for _, v in pairs(admins) do
v = string.lower(v)
if string.sub(v, 1, #name) == name then
--If the first part of the admin name is equal to the name given then it returns that admin is true
admin = true
end
end
return admin
--If there is someone that matches it then it will return they are an admin
end
--Blah blah blah
if getPermission(name) then
--blah
end
This discussion is preserved as it was. New answers, comments, and votes are closed.
2 comments
So best would be the number that matches in the table ? list[1]
I think you overdid it here, Blue. xD