How to make this work

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1 
local distance = 55

local secDictionary = {
 	{distance > 0} == 1,
	{distance > 41} == 2,
	{distance > 90} == 3,
	{distance > 260} == 4,
}

print(secDictionary[distance])

lets say distance is = 30, that would mean it is larger than 0, but not large enough to be larger than 41 so it should print 1 because that is designed number for that distance. I want to make it without for i loops, nor more than 1 if statement, just using table[x} For some reason it prints nil, can somebody help out?

2 Likes
2

You should do something like this:

	--
	local index = 3
	local color
	--
	local colors = {
		[Color3.fromRGB(60, 45, 0)] = {
			["minimum"] = 1,
			["maximum"] = 3,
		},
		[Color3.fromRGB(0, 80, 0)] = {
			["minimum"] = 4,
			["maximum"] = maxIndex,
		}
	}
	--
	for possibleColor, data in colors do
		if index >= data.minimum and index <= data.maximum then
			color = possibleColor
		end
	end
	--

(In your method, 55 is greater than 0 so it would be 1, but its also great than 41 so it would also be 2. It just wont work with 1 variable, so you need a dictionary with 2)

3 
local distance = 30

local thresholds = {
	{limit = 0, value = 1},
	{limit = 41, value = 2},
	{limit = 90, value = 3},
	{limit = 260, value = 4},
}

local result
for i = #thresholds, 1, -1 do
	if distance > thresholds[i].limit then
		result = thresholds[i].value
		break
	end
end

print(result)

Data is data, logic is logic; it’s best to keep them separated. Here, you’re mixing data with logic.

4

yea thats what i wanted, its bigger than 41 so it will be 2 because nuimbers 41+ are asigned to “2”, also is there a way to do that with no for i,v loops? thats what im looking for

5 
local distance = 55

local thresholds = {
	{limit = 0, value = 1},
	{limit = 41, value = 2},
	{limit = 90, value = 3},
	{limit = 260, value = 4},
}

local value = thresholds[1].value
if distance > thresholds[2].limit then value = thresholds[2].value end
if distance > thresholds[3].limit then value = thresholds[3].value end
if distance > thresholds[4].limit then value = thresholds[4].value end

print(value)

Easy to adjust the logic with generic data.

6

There are only two real solutions to this problem, which is doing it procedurally (using a loop, as shown before) or using a pre-computed look-up table. The alternative is to generate a table with each possible input value,

-- Setup

-- List of your desired lower limits and values as pairs
local _secDictionary = {
	{0, 1},
	{41, 2},
	{90, 3},
	{260, 4}
}

 -- Sequential array containing each possible key in the range, assuming integer keys
local secDictionary = table.create(_secDictionary[#_secDictionary][1])

-- Setting keys 1 to #secDictionary in secDictionary with the corresponding value
for k, v in ipairs(_secDictionary) do
	local val = v[2] 
	local stop = (_secDictionary[k + 1] or v)[1] -- Get the next limit
	for i = v[1], stop do
		secDictionary[i] = val
	end
end

-- Usage

local distance = 55

-- If you have a known upper bound, you can exclude this check
if distance <= #secDictionary then
	print(secDictionary[distance])
else
	print(secDictionary[#secDictionary])
end

The problem is similar to weighted random distribution. You might want to look into solutions for those to see if any are relevant, too.

There is no feature in Lua to perform ranges like this. Use a function and it’ll be a one-liner regardless of complexity.