Problem with table sorting

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#1

G’day, I am trying to make a script where players’ score is ordered in a table, the problem is I keep getting this error: “invalid order function for sorting” This is a little strange to me, because it is inconsistent. Sometimes it errors and sometimes it doesn’t, why is this?

local score = (player.RaceInfo.TotalCheckpoints.Value + sectionPercentage) * player.Laps.Value
local newTable = {}
table.insert(newTable, score)
table.insert(newTable, player.Name)
table.insert(RaceTable, newTable)

table.sort(RaceTable, function(a, b)
	return a[1] >= b[1]  
end)
#2

i don’t think you can sort with >=, just > and <

#3

But I have been sorting forever with that, plus it works most of the time, just like it errors once every session, which is strange.

#4

i think that is because the table values are the same and that causes them to error. it fails to sort when it can’t compare them that way.

#5

Okay first thing, I got this answer from here:

and here:

I looked around because this question interested me. This behaviour is happening because if comp(element1, element2) is false, not comp(element2, element1) should also be false. These are two elements we must consider:

First off,

local function sortFuncInvalid(one,two)
	return one[1] >= two[1] -- if this is true, then one[1] > two[1] also has to be true– but if we pass equal values, it won't be true.
	-- by putting this, you're saying 1 > 1 which as stated above, one cannot be greater than one, therefore this function should return false but it doesn't.
	-- 1 is not greater than 1. by doing this, you're saying that one is equal to or greater than two, which is true, but by that same logic, you're saying two is smaller one 1 which is not true
end

Second,

local function sortFuncValid(one,two)
	return one[1] < two[1] -- now, by using this, we are checking that one is smaller than two, which is false because 1 is not smaller than 1, so this would return true.
	-- by the same logic used above, by checking that one < two, we are also checking that two >= one, which is true because 1 is equal or greater than 1
	-- now, if we called the same function with the order swapped, then we use the `not` operator on it, we are checking if one >= two, which once again, is true. 1 is equal to or greater than 1
end

Consensus: 1 is never greater than 1, but 1 is always smaller than or equal to 1.

#6

So what you are saying is to use the 2nd option, as the logic works that way for that function.

#7

Yeah, exactly. I honestly had a hard time understanding it and it really is a difficult concept to understand but table.sort’s source code uses loops (while something do), so checking if the function is valid is essential in preventing a stack overflow.

#8

Ah ok thank you but is it supposed to order them backwards now? Because that is what it is doing.

#9

I don’t think so, try switching the sign (< to > or vice versa).