Lua Table Sorting 2 compares

Question

I have gone over as many of the answers here as I could find that had titles I considered near enough to my problem to look into. I haven't seen anyone having my exact issue, so I'm asking a question I hope is just me being ignorant to a simple fact.

I'm trying to code a table that records HP (int) and the distance (boolean) and then sort by HP with only the ones in Range near the top.

local tableTest = {
    {hp = 64, range = true, name="Frank"},
    {hp = 100, range = true, name="Joe"},
    {hp = 2, range = false, name="Jim"},
    {hp = 76, range = true, name="Tim"},
    {hp = 17, range = false, name="Jill"},
    {hp = 16, range = true, name="Phillip"},
}

-- Sort by HP and Range to find lowest Unit in Range.
table.sort(tableTest, function(x,y) return x.hp < y.hp and x.range end)

for i=1, #tableTest do print(tableTest[i].name, tableTest[i].hp) end

The output for this is:

Phillip 16
Jim     2
Frank   64
Jill    17
Tim     76
Joe     100

The output I was expecting from this would be:

Phillip 16
Frank   64
Tim     76
Joe     100
Jim     2
Jill    17

I pray this is just a misunderstanding on my part of how the table.sort works with multiple checks like this (I assumed it was closer to how you declare a variable like this).

edit Additional information - If I change the order of where the range=false indexes are located in the table, the output changes as well (still incorrect). The values just sort themselves into different indexes after the sort.

score 4 · Accepted Answer · answered Sep 26 '13 at 00:57

4

According to your description, your order function needs to compare range first, then compare hp.

table.sort(tableTest, function(x,y) 
                          if x.range and y.range then return x.hp < y.hp 
                          elseif x.range then return true
                          elseif y.range then return false
                          else return x.hp < y.hp end
                      end)

Maybe there is some shorter version, but this one sure works and the logic is clear.

answered Sep 26 '13 at 00:57

Yu Hao

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ty ty! I think I kinda of understand what I did wrong. I was thinking in terms of return true's only. So if I'm to understand this properly I just want to make sure I'm reading this right. First we check that both ranges are true, then we sort by HP. If both ranges are not true, we will then see if the x has a true range, then return true, else we see if y has the range and return false. If both are false, then we sort the ones outside of range by HP. Where I failed in my tests was the elseif y.range return false. We do this to help y value be sorted properly correcty? – Bubba911 Sep 26 '13 at 01:26
@Bubba911 In that case, `x.range` is `false` while `y.range` is `true`, thus `x` should be after `y` no matter what value of `hp`. – Yu Hao Sep 26 '13 at 02:51
Ty for clarifying that :) My mind was still wrapping itself around it. After verbally saying it all aloud to myself a couple times, what you said is what I came up with to (after posting). So it's great hearing that my understanding it correct! Thank You again, hope you have a great day you fine sir. – Bubba911 Sep 26 '13 at 03:46

score 0 · Answer 2 · answered Sep 27 '13 at 00:20

You already got an answer to this question but I think it's worth adding another one here that covers how you can reason about this logic easier. The ideas presented here are really language agnostic.

The purpose of providing a comparison function is really to answer one question: Should x come before y? Another way to ask the same question is, does x have a higher precedence then y? Often you implement this with the same ordering properties as say the < operator.

So your function should return true only if x definitely precedes y. In your case, you are really sorting by the range field first and if they both happen to be true then use the hp field as a "tie-breaker".

You can construct a truth table here to help you find the most concise way to express the logical condition that yields the behavior you're looking for:

  x  |  y  |  x before y?
-------------------------
  T  |  T  |  x.hp < y.hp
  T  |  F  |  T
  F  |  T  |  F
  F  |  F  |  F

Your original condition x.hp < y.hp and x.range is close but not quite correct for all possible cases.

Above we see that if x is false then the final outcome is also false regardless of what y is. So y is only considered when x is true. Lastly, to avoid the caveat of a falsey condition in the logic short-circuit in lua we'll want x.hp < y.hp to be at the end of the logical expression. Thus, the logical condition you're looking for is:

  return x.range and (not y.range or x.hp < y.hp)

Lua Table Sorting 2 compares

2 Answers2