Min(TSource) Method (IEnumerable(TSource), Func(TSource, Double))

Enumerable.Min(Of TSource) Method (IEnumerable(Of TSource), Func(Of TSource, Double))


Invokes a transform function on each element of a sequence and returns the minimum Double value.

Namespace:   System.Linq
Assembly:  System.Core (in System.Core.dll)

Public Shared Function Min(Of TSource) (
	source As IEnumerable(Of TSource),
	selector As Func(Of TSource, Double)
) As Double


Type: System.Collections.Generic.IEnumerable(Of TSource)

A sequence of values to determine the minimum value of.

Type: System.Func(Of TSource, Double)

A transform function to apply to each element.

Return Value

Type: System.Double

The minimum value in the sequence.

Type Parameters


The type of the elements of source.

Exception Condition

source or selector is null.


source contains no elements.

The Min(Of TSource)(IEnumerable(Of TSource), Func(Of TSource, Double)) method uses the Double implementation of IComparable(Of T) to compare values.

You can apply this method to a sequence of arbitrary values if you provide a function, selector, that projects the members of source into a numeric type, specifically Double.

In Visual Basic query expression syntax, an Aggregate Into Min() clause translates to an invocation of Min.

The following code example demonstrates how to use Min(Of TSource)(IEnumerable(Of TSource), Func(Of TSource, Int32)) to determine the minimum value in a sequence of projected values.


This code example uses an overload of this overloaded method that is different from the specific overload that this topic describes. To extend the example to this topic, change the body of the selector function.

Structure Pet
    Public Name As String
    Public Age As Integer
End Structure

Sub MinEx4()
    ' Create an array of Pet objects.
    Dim pets() As Pet = {New Pet With {.Name = "Barley", .Age = 8}, _
                         New Pet With {.Name = "Boots", .Age = 4}, _
                         New Pet With {.Name = "Whiskers", .Age = 1}}

    ' Find the youngest pet by passing a 
    ' lambda expression to the Min() method.
    Dim min As Integer = pets.Min(Function(pet) pet.Age)

    ' Display the result.
    MsgBox("The youngest pet is age " & min)
End Sub

' This code produces the following output:
' The youngest pet is age 1

Universal Windows Platform
Available since 4.5
.NET Framework
Available since 3.5
Portable Class Library
Supported in: portable .NET platforms
Available since 2.0
Windows Phone Silverlight
Available since 7.0
Windows Phone
Available since 8.1
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