How can I find the words consist of the letters inside the keyword in Lua?

Question

For example, I have a keyword \"abandoned\" and I want to find the words that contains letters of this keyword such as \"done\", \"abandon\", band\", from the arrays I stored t

Answer 1

Try this:

W={"done", "abandon", "band"}
for k,w in pairs(W) do
    W[w]=true
end

function findwords(s)
    for i=1,#s do
        for j=i+1,#s do
            local w=s:sub(i,j)
            if W[w] then print(w) end
        end
    end
end

findwords("abandoned")

If you don't have an array of words, you can load a dictionary:

for w in io.lines("/usr/share/dict/words") do
    W[w]=true
end

Answer 2

For example, I have a keyword "abandoned" and I want to find the words that contains letters of this keyword such as "done", "abandon", band", from the arrays I stored those words. How can I search it?

You can simply use the keyword as a regular expression (aka "pattern" in Lua), using it's letters as a set, for instance ('^[%s]+$'):format('abandoned'):match('done').

local words = {'done','abandon','band','bane','dane','danger','rand','bade','rand'}
local keyword = 'abandoned'

-- convert keyword to a pattern and match it against each word
local pattern = string.format('^[%s]+$', keyword)
for i,word in ipairs(words) do
    local matches = word:match(pattern)
    print(word, matches and 'matches' or 'does not match')
end

Output:

done    matches
abandon matches
band    matches
bane    matches
dane    matches
danger  does not match
rand    does not match
bade    matches
rand    does not match

Answer 3

Run the loop on your arrays and use string.find to check against this long word.

for idx = 1, #stored_words do
   local word = stored_words[idx]
   if string.find(long_word, word, 1, true) then
      print(word .. " matches part of " .. long_word)
   end
end