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Best way to order values in size?
If I had 8 values is there a way I can easily order them in size order, without lots of lines of code? Example with 3 values
Values = [1, 2, 3]
For i in Values do
if Values[1] > Values[2] > Values[3] then
print (Values[1].." is bigger than "..Values[2].." which is bigger than "..Values[3])
if Values[2] > Values[3] > Values[1] then
print (Values[2].." is bigger than "..Values[3].." which is bigger than "..Values[1])
-- etc
As you can see this is inefficient, is there a more efficient way, considering I require 8 values
1 answer
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The best, most efficient way to do this would be using the table.sort function.
So, you have a table full of numbers and you want to sort them from least to greatest?
local values = {5,3,7,2,1,8}
table.sort(values) --Sort them
print(table.concat(values," ")) --Print the whole table, with a space between each index.
This code would print; 1 2 3 5 7 8
--EDIT--
table.concat is a method that takes 2 arguments. The first argument should be the table, and the the second should be the intersection between each index of the table.
Example;
local tab = {"I","Really","Like","Roblox"}
local sentence = table.concat(tab," ")
--[[First argument is the table, 'tab'. The second is what goes in between each index, " ", or a space.]]
print(sentence) --> I Really Like Roblox
--[[Alternatively, if I were to make the second argument ", " then it would put a comma in between each index.]]
local sentence = table.concat(tab,", ")
print(sentence) --> I, Really, Like, Roblox
More info about table.concat here.
Also, you can use the unpack function.
