Get max value from an dictionary in relation to a value

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#1

I have a dictionary that looks like this:

local Items = {
    ["Item1"] = 500, 
    ["Item2"] = 1500, 
    ["Item3"] = 5000, 
}

return Items

and a player has 4000 dollars then it would automatically detect the highest valued item it can afford (in this context it would be “Item2”.

What would be my best way to do this

#2 
local highestValue = 0
local highestItem 
for item, cost in ipairs(Items) do
    if cost > highestValue and cost <= money then
        highestValue = cost
        highestItem = item
    end
end
#3

This isn’t detecting any items for me.

	local highestValue = 3750000
	local highestItem 
	for item, cost in ipairs(Items) do
		if cost > highestValue and cost <= Player:WaitForChild("leaderstats").Cash then
			highestValue = cost
			highestItem = item
			print(highestItem)
		end
	end

It isn’t printing anything so I think that it hasn’t found any items

#4

Because highestValue is set to a number higher than any of the costs in the dictionary.

#5

I made highest value the value of the highest item in my current dictionary. The dictionary in the OG post was just a example one.

#6

The check doesn’t consider if the costs are equal. I would just set highestValue to 0 so that it works for all costs.

#7

Even with it set to 0 I don’t get anything printed.

#8 
local function GetHighestFromNumber(table,number)
   local Highest = 0
   for _,v in pairs(table) do
      if Highest < v and number >= v then Highest = v end
   end

   return Highest
end

if u want to return index to then

local function GetHighestFromNumber(table,number)
   local Highest,index = 0,0
   for i,v in pairs(table) do
      if Highest < v and number >= v then Highest = v index = i end
   end

   return Highest, index
end
1 Like
#9 
local function countDictionary(dictionary)
	local count = 0
	for key, value in next, dictionary do
		count += 1
		if not next(dictionary, key) then
			return count
		end
	end
end

Not what you asked for but here’s a dictionary count function.

#10 
local function convertToArray(dictionary)
	local array = {}
	for key, value in next, dictionary do
		table.insert(array, {key, value})
	end
	return array
end

local function sortDictionary(dictionary)
	local array = convertToArray(dictionary)
	table.sort(array, function(left, right)
		if left[2] > right[2] then
			return left
		end
	end)
	return array
end

local dict = sortDictionary({["a"] = 1, ["b"] = 2, ["c"] = 3})
for _, data in ipairs(dict) do
	print(data[1], data[2])
end

--[[
output:
c 3
b 2
a 1
]]

and here’s a dictionary sorting function which doesn’t just return the item/entry with the largest value but it returns all items/entries in descending order of their values (largest to smallest).

#11

One issue with this is it won’t be able to sort negative numbers as the variable “Highest” starts at 0.

local function getLargestValue(dictionary)
	local index, largestValue = nil, -math.huge
	for key, value in next, dictionary do
		if value > largestValue then
			largestValue = value
			index = key
		end
	end
	return index, largestValue
end

local function getSmallestValue(dictionary)
	local index, smallestValue = nil, math.huge
	for key, value in next, dictionary do
		if value < smallestValue then
			smallestValue = value
			index = key
		end
	end
	return index, smallestValue
end

local dict = {["a"] = 1, ["b"] = 2, ["c"] = 3}

local index, largestValue = getLargestValue(dict)
print(index, largestValue) --a 3
local index, smallestValue = getSmallestValue(dict)
print(index, smallestValue) --c 1

Here we compare against -math.huge/math.huge to circumvent that issue.

1 Like
#12

we are talking about Shop so this wont be needed

#13

A general implementation is more versatile.

1 Like
#14

Fair, can be more customizable