Creating C# SortedSet from existing collection

Question

I have an existing HashSet with 6 items:{ Value: 3, Occurrence: 1 },{ Value: 1, Occurrence: 2 },{ Value: 4, Occurrence: 2 },{ Value: 5, Occurrence: 1 },{ Value: 2, Occurrence: 1 },{ Value: 6, Occurrence: 1 }

Element class:

internal class Element
{
    public Element(int value)
    {
         this.Value = value;
          this.Occurrence = 1;
    }

    public int Value { get; set; }

    public int Occurrence { get; set; }
}

How I want to create a SortedSet from items of this hash set like this:

var sortedSet = new SortedSet<Element>(hashSet.AsEnumerable(), new SortedSetComparer());

SortedSetComparer:

 internal class SortedSetComparer : IComparer<Element>
 {
     public int Compare(Element x, Element y)
     {
         if (x != null && y != null)
         {
            if (x.Occurrence > y.Occurrence)
            {
                return 1;
            }
            if (y.Occurrence > x.Occurrence)
            {
                return -1;
            }

            return 0;
        }

        return 0;
    }
}

But in debug I see that only 2 first elements got into sorted set: {Value: 3, Occurrence: 1} and {Value: 1, Occurrence: 2}

What am I doing wrong?

You are not comparing the Value is your main issue. You are effectively saying { Value: 1, Occurrence: 2 } and { Value: 4, Occurrence: 2 } are the same because that is what 0 signifies. — mjwills
– mjwills, Commented Jun 21, 2019 at 14:12
I want them to be sorted by Occurrence not Value. Anyway shouldn't they all go to sorted set regardless of underlying comparer? — SimonD
– SimonD, Commented Jun 21, 2019 at 14:15
'A set has no duplicates' - I didn't know that, my fault.. Could you post your answer and I'll accept it? — SimonD
– SimonD, Commented Jun 21, 2019 at 14:19
The .NET framework doesn't contain a sortable collection that allows duplicate keys out of the box. You will have to use tricks if you need one. — Theodor Zoulias
– Theodor Zoulias, Commented Jun 21, 2019 at 14:47

mjwills · Accepted Answer · 2019-06-21 14:21:39Z

2

As per the docs (and by definition):

Duplicate elements are not allowed.

Since in your comparison method you are returning 0 if two objects have the same Occurrence (but different Value), then the set believes those two objects are equal. The net effect - it adds the first item for each Occurrence value and ignores the rest.

To solve this issue, your comparison must compare Occurrence and then compare Value as well. 0 should be returned only if both Occurrence and Value are equal.

answered Jun 21, 2019 at 14:21

mjwills

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