java collection sort issue

Question

i use simple comperator and get exception and don't know what to do

this is how i call:

try {
   Collections.sort(this.closePositions, new PositionComperator());
}
catch(Exception e) {
   e.printStackTrace();
}

this is the comperator:

  public class PositionComperator implements Comparator<DataResponse> {

    @Override
    public int compare( DataResponse pos1, DataResponse pos2) {

        if (pos1.openTime >= pos2.openTime) {
            return 1;
        } 
        else {
            return -1;
        }// returning 0 would merge keys

    }

   }

this is the exception:

java.lang.IllegalArgumentException: Comparison method violates its general contract!
at java.util.TimSort.mergeLo(Unknown Source)
at java.util.TimSort.mergeAt(Unknown Source)
at java.util.TimSort.mergeCollapse(Unknown Source)
at java.util.TimSort.sort(Unknown Source)
at java.util.TimSort.sort(Unknown Source)
at java.util.Arrays.sort(Unknown Source)
at java.util.Collections.sort(Unknown Source)
at GTTask.RefreshIdentityHistory.call(RefreshIdentityHistory.java:59)
at GTTask.RefreshIdentityHistory.call(RefreshIdentityHistory.java:1)
at java.util.concurrent.FutureTask$Sync.innerRun(Unknown Source)
at java.util.concurrent.FutureTask.run(Unknown Source)
at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source)
at java.lang.Thread.run(Unknown Source)

sort can't cause elements to get merged. Though if you were to use a Set, it would be a different story. — Bernhard Barker
– Bernhard Barker, Commented Jul 31, 2013 at 8:14

Jon Skeet · Accepted Answer · 2013-07-31 08:13:45Z

2

If two values x and y have the same openTime, then compare(x, y) and compare(y, x) will both return 1, which violates the contract of compare:

The implementor must ensure that sgn(compare(x, y)) == -sgn(compare(y, x)) for all x and y.

You haven't ensured that.

You need to consider what you want to happen when the openTime values are the same - either return 0, or have some consistent notion of which value should come before the other. Do you have some secondary comparison you could perform, for example?

answered Jul 31, 2013 at 8:13

Jon Skeet

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1 Comment

Jon Skeet Over a year ago

@user502967: No, for exactly the reason I gave: if the two values are equal, you'll get 1 when comparing them both ways. Look at the contract for compare.

Knubo · Accepted Answer · 2013-07-31 08:16:03Z

1

The reason you get this error is that when it sorted two items they changed order. You should also include the case where it is equal.

Preferably do:

return po1.openTime - pos2.opentime;

Or do

if (pos1.openTime > pos2.openTime) {
    return 1;
} 
else if (pos1.openTime < pos2.openTime) {
    return -1;
} else {
    return 0;
}

answered Jul 31, 2013 at 8:16

Knubo

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2 Comments

user502967 Over a year ago

what does "return 0" will cause?

Knubo Over a year ago

This means that they are equal and they will not be sorted. Have a look at the java doc for the compare method.

user2583040 · Accepted Answer · 2013-07-31 08:20:43Z

1

you can use treeSet. İt is sort it for you. And there is compare method. for example

TreeSet<Double> sortedSet = new TreeSet<Double>();

for example compare it

TreeSet<Double> set = new TreeSet<Rock>(new Comparator<Double>()
public int compare(Double a, Double b){
                return a.value - b.value;
            }
        }

edited Jul 31, 2013 at 8:20

answered Jul 31, 2013 at 8:14

user2583040

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2 Comments

user502967 Over a year ago

why to use tree sort on list?

user2583040 Over a year ago

you can look this topic why you should use treeset. Treeset is directly sort and compare it you dont need another method and you can see more easy. stackoverflow.com/questions/1463284/hashset-vs-treeset

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