Enumerable.Concat<TSource>(IEnumerable<TSource>, IEnumerable<TSource>) メソッド

定義

2 つのシーケンスを連結します。Concatenates two sequences.

public:
generic <typename TSource>
[System::Runtime::CompilerServices::Extension]
 static System::Collections::Generic::IEnumerable<TSource> ^ Concat(System::Collections::Generic::IEnumerable<TSource> ^ first, System::Collections::Generic::IEnumerable<TSource> ^ second);
public static System.Collections.Generic.IEnumerable<TSource> Concat<TSource> (this System.Collections.Generic.IEnumerable<TSource> first, System.Collections.Generic.IEnumerable<TSource> second);
static member Concat : seq<'Source> * seq<'Source> -> seq<'Source>
<Extension()>
Public Function Concat(Of TSource) (first As IEnumerable(Of TSource), second As IEnumerable(Of TSource)) As IEnumerable(Of TSource)

型パラメーター

TSource

入力シーケンスの要素の型。The type of the elements of the input sequences.

パラメーター

first
IEnumerable<TSource>

連結する最初のシーケンス。The first sequence to concatenate.

second
IEnumerable<TSource>

最初のシーケンスに連結するシーケンス。The sequence to concatenate to the first sequence.

戻り値

IEnumerable<TSource>

2 つの入力シーケンスの連結された要素が格納されている IEnumerable<T>An IEnumerable<T> that contains the concatenated elements of the two input sequences.

例外

first または secondnull です。first or second is null.

次のコード例では、Concat<TSource>(IEnumerable<TSource>, IEnumerable<TSource>) を使用して2つのシーケンスを連結する方法を示します。The following code example demonstrates how to use Concat<TSource>(IEnumerable<TSource>, IEnumerable<TSource>) to concatenate two sequences.

class Pet
{
    public string Name { get; set; }
    public int Age { get; set; }
}

static Pet[] GetCats()
{
    Pet[] cats = { new Pet { Name="Barley", Age=8 },
                   new Pet { Name="Boots", Age=4 },
                   new Pet { Name="Whiskers", Age=1 } };
    return cats;
}

static Pet[] GetDogs()
{
    Pet[] dogs = { new Pet { Name="Bounder", Age=3 },
                   new Pet { Name="Snoopy", Age=14 },
                   new Pet { Name="Fido", Age=9 } };
    return dogs;
}

public static void ConcatEx1()
{
    Pet[] cats = GetCats();
    Pet[] dogs = GetDogs();

    IEnumerable<string> query =
        cats.Select(cat => cat.Name).Concat(dogs.Select(dog => dog.Name));

    foreach (string name in query)
    {
        Console.WriteLine(name);
    }
}

// This code produces the following output:
//
// Barley
// Boots
// Whiskers
// Bounder
// Snoopy
// Fido
Structure Pet
    Public Name As String
    Public Age As Integer
End Structure

' Returns an array of Pet objects.
Function GetCats() As Pet()
    Dim cats() As Pet = {New Pet With {.Name = "Barley", .Age = 8},
                 New Pet With {.Name = "Boots", .Age = 4},
                 New Pet With {.Name = "Whiskers", .Age = 1}}

    Return cats
End Function

' Returns an array of Pet objects.
Function GetDogs() As Pet()
    Dim dogs() As Pet = {New Pet With {.Name = "Bounder", .Age = 3},
                 New Pet With {.Name = "Snoopy", .Age = 14},
                 New Pet With {.Name = "Fido", .Age = 9}}
    Return dogs
End Function

Sub ConcatEx1()
    ' Create two arrays of Pet objects.
    Dim cats() As Pet = GetCats()
    Dim dogs() As Pet = GetDogs()

    ' Project the Name of each cat and concatenate
    ' the collection of cat name strings with a collection
    ' of dog name strings.
    Dim query As IEnumerable(Of String) =
cats _
.Select(Function(cat) cat.Name) _
.Concat(dogs.Select(Function(dog) dog.Name))

    Dim output As New System.Text.StringBuilder
    For Each name As String In query
        output.AppendLine(name)
    Next

    ' Display the output.
    MsgBox(output.ToString())
End Sub

' This code produces the following output:
'
' Barley
' Boots
' Whiskers
' Bounder
' Snoopy
' Fido

2つのシーケンスを連結する別の方法として、シーケンスのコレクション (配列など) を構築し、SelectMany メソッドを適用して id セレクター関数を渡す方法があります。An alternative way of concatenating two sequences is to construct a collection, for example an array, of sequences and then apply the SelectMany method, passing it the identity selector function. 次の例では、SelectMany の使用方法を示します。The following example demonstrates this use of SelectMany.

Pet[] cats = GetCats();
Pet[] dogs = GetDogs();

IEnumerable<string> query =
    new[] { cats.Select(cat => cat.Name), dogs.Select(dog => dog.Name) }
    .SelectMany(name => name);

foreach (string name in query)
{
    Console.WriteLine(name);
}

// This code produces the following output:
//
// Barley
// Boots
// Whiskers
// Bounder
// Snoopy
// Fido
    ' Create two arrays of Pet objects.
    Dim cats() As Pet = GetCats()
    Dim dogs() As Pet = GetDogs()

    ' Create an IEnumerable collection that contains two elements.
    ' Each element is an array of Pet objects.
    Dim animals() As IEnumerable(Of Pet) = {cats, dogs}

    Dim query As IEnumerable(Of String) =
(animals.SelectMany(Function(pets) _
                        pets.Select(Function(pet) pet.Name)))

    Dim output As New System.Text.StringBuilder
    For Each name As String In query
        output.AppendLine(name)
    Next

    ' Display the output.
    MsgBox(output.ToString())

    ' This code produces the following output:
    '
    ' Barley
    ' Boots
    ' Whiskers
    ' Bounder
    ' Snoopy
    ' Fido

注釈

このメソッドは、遅延実行を使用して実装されます。This method is implemented by using deferred execution. イミディエイトの戻り値は、アクションを実行するために必要なすべての情報を格納するオブジェクトです。The immediate return value is an object that stores all the information that is required to perform the action. か、呼び出すことによって、オブジェクトが列挙されるまで、このメソッドによって表されるクエリは実行されません、GetEnumeratorメソッドを使用して直接またはforeachVisual C# またはFor EachVisual Basic で。The query represented by this method is not executed until the object is enumerated either by calling its GetEnumerator method directly or by using foreach in Visual C# or For Each in Visual Basic.

@No__t-0 メソッドは、Union メソッドとは異なります。これは、Concat<TSource>(IEnumerable<TSource>, IEnumerable<TSource>) メソッドが入力シーケンス内のすべての元の要素を返すためです。The Concat<TSource>(IEnumerable<TSource>, IEnumerable<TSource>) method differs from the Union method because the Concat<TSource>(IEnumerable<TSource>, IEnumerable<TSource>) method returns all the original elements in the input sequences. @No__t-0 メソッドは、一意の要素のみを返します。The Union method returns only unique elements.

適用対象