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Brian Graves
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Brian Graves asked on 20 May 2010, 01:41 AM
I am using the SumFunction for my Column AggregateFunctions.  However, when I use the CountFunction it simply returns the number of rows in my grid.  I was hoping that it would return the sum of all UNIQUE values in my Column.  Is there a way to do a DISTINCT count using the CountFunction?  If not, how can I programmatically bind a function to my column that will return the number of unique values?

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Pavel Pavlov
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answered on 21 May 2010, 08:52 AM
Hello Brian Graves,

The attached project demonstrates how to create a custom aggregate function and use it in RadGridView.
For your needs you will need to modify  MyAggregateMethod. Within the method you have access to the items in RadGridView and may write a few lines of code to return the distinct total.
Let me know in case you need further assistance.

Sincerely yours,
Pavel Pavlov
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Brian Graves
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answered on 22 May 2010, 01:03 AM
Thanks...this worked. However, I found that I needed to create a different MyAggregationMethod for every single field in my grid (that I need to do a distinct count on).  Is there any way I can pass a string from the MyAggregateFunction class to the MyAggregateMethod so I know which column that I want to do a distinct sum on?
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Pavel Pavlov
Telerik team
answered on 25 May 2010, 01:12 PM
Hello Brian Graves,

I am afraid that it is not possible to pass parameter to the method this way . However I think we can find an alternative way - we can make a custom EnumerableSelectorAggregateFunction instead. This one has the SourceFiled property.

I will prepare a small sample , but I will need to know the data type of the property you will make distinct count on.

All the best,
Pavel Pavlov
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Brian Graves
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answered on 25 May 2010, 04:22 PM
There isn't a single data type.  I could be doing a distinct count on a string column, a date column, dollar amount, or even an integer column...it depends what I want to do a distinct count on.  Anyway, I figured out how to do this using the EnumerableSelectorAggregateFunction you've suggested:

 

    public class CountDistinctFunction : EnumerableSelectorAggregateFunction
    {
        protected override string AggregateMethodName
        {
            get
            {
                return "CountDistinct";
            }
        }

        protected override Type ExtensionMethodsType
        {
            get
            {
                return typeof(CountDistinctAggregates);
            }
        }
    }

    public static class CountDistinctAggregates
    {
        public static int CountDistinct<TSource>(IEnumerable<TSource> source, Func<TSource, float> selector)
        {
            if (!source.Any<TSource>())
                return 0;
            else
                return source.Select(selector).Distinct().Count();
        }

        public static int CountDistinct<TSource, TResult>(IEnumerable<TSource> source, Func<TSource, TResult> selector)
        {
            if (!source.Any<TSource>())
                return 0;
            else
            {
                IEnumerable<string> values = from i in source select Convert.ToString(selector(i)).ToLower();
                var result = values.Distinct().Count();
                return result;
            }
        }

        public static int CountDistinct<TSource>(IEnumerable<TSource> source, Func<TSource, int> selector)
        {
            if (!source.Any<TSource>())
                return 0;
            else
                return source.Select(selector).Distinct().Count();
        }
    }

 

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Brian Graves
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answered on 26 May 2010, 04:13 PM
Oh...forgot to mention.  Here's how you add a CountDistinctFunction to your column:

col.AggregateFunctions.Add(
new CountDistinctFunction { SourceField = columnName });

 

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Pavel Pavlov
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