Scripting Helpers is winding down operations and is now read-only. More info→
Log in to vote
0
How can i sort the second value of a table?
I need the table Ores to be sorted but the first value is an object and i need it to be sorted by the Magnitude wich is the second value
for _, Ore in pairs(SpotObjects:GetChildren()) do
if Ore:IsA("BasePart") then
if Ore.Lvl1.Transparency == 0 then
OreMagnitude = (HumanoidRootPart.Position - Ore.Position).Magnitude
Ores[Ore] = OreMagnitude
end
end
end
table.sort(Ores)
for OresObject, i in pairs(Ores) do
print(OresObject, i)
end
When i do print its not printing it in order
1 answer
Log in to vote
2
table.sort does not work on dictionaries; infact it is impossible to sort dictionaries. If your goal's to end up with an ore array sorted close to far, I have a better script:
local RootPart = --?
local ores = {}
--create arbitrary array of ores
for _,ore in ipairs(SpotObjects:GetChildren()) do
if ore:IsA("BasePart") and ore.Lvl1.Transparency==0 then
ores[#ores+1]=ore
end
end
--custom sort; sort() takes an optional function which manually compares values
--think of the default function as: function(a,b) return a < b end
table.sort(ores,function(ore1,ore2)
--comparing magnitudes will sort by distance
local mag1 = (ore1.Position-RootPart.Position).magnitude
local mag2 = (ore2.Position-RootPart.Position).magnitude
--you can flip the "<" to sort greatest-to-least
return mag1 < mag2
end)
print(unpack(ores))
might be typos but I hope you get it. (pls dont simply ignore like some ppl e.e)