Queryable.Distinct<TSource> Method (IQueryable<TSource>)
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Returns distinct elements from a sequence by using the default equality comparer to compare values.
Namespace: System.Linq
Assembly: System.Core (in System.Core.dll)
Syntax
'Declaration
<ExtensionAttribute> _
Public Shared Function Distinct(Of TSource) ( _
source As IQueryable(Of TSource) _
) As IQueryable(Of TSource)
public static IQueryable<TSource> Distinct<TSource>(
this IQueryable<TSource> source
)
Type Parameters
- TSource
The type of the elements of source.
Parameters
- source
Type: System.Linq.IQueryable<TSource>
The IQueryable<T> to remove duplicates from.
Return Value
Type: System.Linq.IQueryable<TSource>
An IQueryable<T> that contains distinct elements from source.
Usage Note
In Visual Basic and C#, you can call this method as an instance method on any object of type IQueryable<TSource>. When you use instance method syntax to call this method, omit the first parameter.
Exceptions
Exception | Condition |
---|---|
ArgumentNullException | source is nulla null reference (Nothing in Visual Basic). |
Remarks
The Distinct<TSource>(IQueryable<TSource>) method generates a MethodCallExpression that represents calling Distinct<TSource>(IQueryable<TSource>) itself as a constructed generic method. It then passes the MethodCallExpression to the CreateQuery<TElement>(Expression) method of the IQueryProvider represented by the Provider property of the source parameter.
The query behavior that occurs as a result of executing an expression tree that represents calling Distinct<TSource>(IQueryable<TSource>) depends on the implementation of the type of the source parameter. The expected behavior is that it returns an unordered sequence of the unique items in source.
Examples
The following code example demonstrates how to use Distinct<TSource>(IQueryable<TSource>) to return distinct elements from a sequence.
Dim ages As List(Of Integer) = New List(Of Integer)(New Integer() {21, 46, 46, 55, 17, 21, 55, 55})
Dim distinctAges As IEnumerable(Of Integer) = ages.AsQueryable().Distinct()
Dim output As New System.Text.StringBuilder
output.AppendLine("Distinct ages:")
For Each age As Integer In distinctAges
output.AppendLine(age)
Next
' Display the output.
outputBlock.Text &= output.ToString() & vbCrLf
' This code produces the following output:
'
' Distinct(ages)
' 21
' 46
' 55
' 17
List<int> ages = new List<int> { 21, 46, 46, 55, 17, 21, 55, 55 };
IEnumerable<int> distinctAges = ages.AsQueryable().Distinct();
outputBlock.Text += "Distinct ages:" + "\n";
foreach (int age in distinctAges)
outputBlock.Text += age + "\n";
/*
This code produces the following output:
Distinct ages:
21
46
55
17
*/
Version Information
Silverlight
Supported in: 5, 4, 3
Silverlight for Windows Phone
Supported in: Windows Phone OS 7.1
Platforms
For a list of the operating systems and browsers that are supported by Silverlight, see Supported Operating Systems and Browsers.