string algo DO ALL MATH

-- 3rd attempt in string algo math thing

-- the DO ALL MATH algorithm actually works, however it thinks all decimals are numbers which will be fixed later.

str = "10000/333333333 is probably a much larger number than 0.9999999999+0.0000000001, but I'm not really too sure. Also 23*3 is a cool number."

function isnumber(x)
    x = tostring(x)
    local numbs = {"1","2","3","4","5","6","7","8","9","0","."}
    for i=1,#numbs do
        if numbs[i] == x then
            return true
        end
    end
    return false
end

function isoperator(x)
    local numbs = {"*","/","+","-"}
    for i=1,#numbs do
        if numbs[i] == x then
            return {true,numbs[i]}
        end
    end
    return {false,nil}
end

function calculate(x,y,z)
    if y == "+" then
        return x+z
    elseif y == "-" then
        return x-z
    elseif y == "*" then
        return x*z
    elseif y == "/" then
        return x/z
    end
end


local num = false
local firstfound = false
local sub1
local sub2
local sub3
local op
local strr = #str

function resetvars()
    num = false
    firstfound = false
    sub1 = nil
    sub2 = nil
    sub3 = nil
    op = nil
end

for i=1,strr do
    local x = string.sub(str,i,i)
    if isnumber(x) == true and num == false then
        if x == "." then
            if isnumber(string.sub(str,i+1,i+1)) == true then
                sub1 = i
                firstfound = true
                num = true
            end
        else
            sub1 = i
            firstfound = true
            num = true
        end
    end
    if isnumber(x) == false and num == true and sub1 ~= nil and sub2 == nil and sub3 == nil and isoperator(x)[1] == true then
        sub2 = i
        op = string.sub(str,sub2,sub2)
    elseif isnumber(x) == false and num == true and sub1 ~= nil and sub2 == nil and sub3 == nil and isoperator(x)[1] == false then
        resetvars()
    end
    if isnumber(x) == false and num == true and sub1 ~= nil and sub2 ~= nil and sub3 == nil then
        if i-sub2 > 0 then
            if x == "." then
                if isnumber(string.sub(str,i+1,i+1)) == true then
                    sub3 = i
                end
            else
                sub3 = i
            end
        end
    end
    if (num == true and sub1 ~= nil and sub2 ~= nil and sub3 ~= nil) then
        local number1 = tonumber(string.sub(str,sub1,sub2-1))
        local number2 = tonumber(string.sub(str,sub2+1,sub3-1))
        local sum = calculate(number1,op,number2)
        local k = #str
        str = string.sub(str,1,sub1-1)..tostring(sum)..string.sub(str,sub3)
        local k2 = k-#str
        i = i-k2
        strr = #str
        resetvars()
    end
end

print(str)