blocks|key|1781640|text|我想你可能在找LINQ的.Concat()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1781641|var+combined+=+foo.Concat(bar).Concat(foobar).Concat(...);|code-block|syntax|javascript|1781642|或者，.Union()将删除重复的元素。|1781643|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/bb302894%2528v=vs.110%2529.aspx^0|C|9|C|9|0|0|0|3|8|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

I think you might be looking for LINQ's <a href="http://msdn.microsoft.com/en-us/library/bb302894%28v=vs.110%29.aspx" rel="noreferrer"><code>.Concat()</code></a>?

<pre><code>var combined = foo.Concat(bar).Concat(foobar).Concat(...);
</code></pre>

Alternatively, <code>.Union()</code> will remove duplicate elements.

blocks|key|1995964|text|当你有一个集合的时候，比如一个List<List<T>>，Enumerable.Aggregate是一种更优雅的将所有列表组合成一个列表的方式：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1995965|var+combined+=+lists.Aggregate((acc,+list)+=>+{+return+acc.Concat(list);+});|code-block|syntax|javascript|1995966|entityMap^0|F|D|T|K|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

For the case when you do have a collection of collections, i.e. a <code>List&lt;List&lt;T&gt;&gt;</code>, <code>Enumerable.Aggregate</code> is a more elegant way to combine all lists into one:

<pre><code>var combined = lists.Aggregate((acc, list) =&gt; { return acc.Concat(list); });
</code></pre>

blocks|key|1781654|text|像这样使用Enumerable.Concat：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1781655|var+combined+=+foo.Concat(bar).Concat(baz)....;|code-block|syntax|javascript|1781656|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/bb302894.aspx^0|5|H|5|H|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Use <a href="http://msdn.microsoft.com/en-us/library/bb302894.aspx" rel="noreferrer"><code>Enumerable.Concat</code></a> like so:

<pre><code>var combined = foo.Concat(bar).Concat(baz)....;
</code></pre>

blocks|key|1781825|text|您可以始终使用Aggregate和Concat结合使用...|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1781826|++++++++var+listOfLists+=+new+List<List<int>>
++++++++{
++++++++++++new+List<int>+{1,+2,+3,+4},
++++++++++++new+List<int>+{5,+6,+7,+8},
++++++++++++new+List<int>+{9,+10}
++++++++};

++++++++IEnumerable<int>+combined+=+new+List<int>();
++++++++combined+=+listOfLists.Aggregate(combined,+(current,+list)+=>+current.Concat(list)).ToList();|code-block|syntax|javascript|1781827|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You could always use Aggregate combined with Concat...

<pre><code> var listOfLists = new List&lt;List&lt;int&gt;&gt;
 {
 new List&lt;int&gt; {1, 2, 3, 4},
 new List&lt;int&gt; {5, 6, 7, 8},
 new List&lt;int&gt; {9, 10}
 };

 IEnumerable&lt;int&gt; combined = new List&lt;int&gt;();
 combined = listOfLists.Aggregate(combined, (current, list) =&gt; current.Concat(list)).ToList();
</code></pre>

blocks|key|1995413|text|使用Enumerable.Concat|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1995414|var+query+=+foo.Concat(bar);|code-block|syntax|javascript|1995415|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/bb302894.aspx^0|2|H|2|H|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Use <a href="http://msdn.microsoft.com/en-us/library/bb302894.aspx" rel="nofollow noreferrer"><code>Enumerable.Concat</code></a>:
<pre><code>var query = foo.Concat(bar);
</code></pre>

blocks|key|1781615|text|您可以按如下方式使用Union：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1781616|var+combined=foo.Union(bar).Union(baz)...|code-block|syntax|javascript|1781617|这将删除相同的元素，因此如果您有这些元素，您可能希望使用Concat。|offset|length|style|CODE|1781618|entityMap^0|0|0|S|6|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You can use Union as follows:

<pre><code>var combined=foo.Union(bar).Union(baz)...
</code></pre>

This will remove identical elements, though, so if you have those, you might want to use <code>Concat</code>, instead.

blocks|key|1996008|text|使用集合初始化器的一些技术--|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1996009|假设这些列表：|1996010|var+list1+=+new+List<int>+{+1,+2,+3+};
var+list2+=+new+List<int>+{+4,+5,+6+};|code-block|syntax|javascript|1996011|带有数组初始化器的SelectMany+(对我来说不是很优雅，但不依赖于任何辅助函数)：|offset|length|style|CODE|1996012|var+combined+=+new+[]{+list1,+list2+}.SelectMany(x+=>+x);|1996013|为允许在List<T>+initializer中使用IEnumerable<T>的Add定义列表扩展|1996014|public+static+class+CollectionExtensions
{
++++public+static+void+Add<T>(this+ICollection<T>+collection,+IEnumerable<T>+items)
++++{
++++++++foreach+(var+item+in+items)+collection.Add(item);
++++}
}|1996015|然后，您可以创建一个包含其他元素的新列表，如下所示(它还允许混合使用单个项目)。|1996016|var+combined+=+new+List<int>+{+list1,+list2,+7,+8+};|1996017|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.microsoft.com/en-us/dotnet/csharp/programming-guide/classes-and-structs/object-and-collection-initializers#collection-initializers^0|0|0|0|9|A|0|0|4|7|Q|E|C|B|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|17|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|18|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|19|8|@$K|1A|L|1B|M|N]]|9|@]|A|$]]|$1|O|3|P|5|F|7|1C|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|1D|8|@$K|1E|L|1F|M|N]|$K|1G|L|1H|M|N]]|9|@$K|1I|L|1J|1|1K]]|A|$]]|$1|S|3|T|5|F|7|1L|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|1M|8|@]|9|@]|A|$]]|$1|W|3|X|5|F|7|1N|8|@]|9|@]|A|$G|H]]|$1|Y|3|-4|5|6|7|1O|8|@]|9|@]|A|$]]]|Z|$10|$5|11|12|13|A|$14|15]]]]

A couple techniques using Collection Initializers --

assuming these lists:

<pre><code>var list1 = new List&lt;int&gt; { 1, 2, 3 };
var list2 = new List&lt;int&gt; { 4, 5, 6 };
</code></pre>

<code>SelectMany</code> with an array initializer (not really that elegant to me, but doesn't rely on any helper function):

<pre><code>var combined = new []{ list1, list2 }.SelectMany(x =&gt; x);
</code></pre>

Define a list extension for Add which allows <code>IEnumerable&lt;T&gt;</code> in the <code>List&lt;T&gt;</code> <a href="https://docs.microsoft.com/en-us/dotnet/csharp/programming-guide/classes-and-structs/object-and-collection-initializers#collection-initializers" rel="nofollow noreferrer">initializer</a>:

<pre><code>public static class CollectionExtensions
{
 public static void Add&lt;T&gt;(this ICollection&lt;T&gt; collection, IEnumerable&lt;T&gt; items)
 {
 foreach (var item in items) collection.Add(item);
 }
}
</code></pre>

Then you can create a new list containing the elements of the others like this (it also allows single items to be mixed in).

<pre><code>var combined = new List&lt;int&gt; { list1, list2, 7, 8 };
</code></pre>

blocks|key|1781797|text|考虑到您是从一堆独立的集合开始的，我认为您的解决方案相当优雅。你得做点什么才能把它们缝起来。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1781798|从语法上讲，在您的组合方法之外创建一个扩展方法会更方便，这将使您可以随时随地使用它。|1781799|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Given that you're starting with a bunch of separate collections, I think your solution is rather elegant. You're going to have to do something to stitch them together.

It would be more convenient syntactically to make an extension method out of your Combine method, which would make it available anywhere you go.

blocks|key|1782211|text|对于任何IEnumerable<IEnumerable<T>>+lists，您所需要的就是：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1782212|var+combined+=+lists.Aggregate((l1,+l2)+=>+l1.Concat(l2));|code-block|syntax|javascript|1782213|这将把lists中的所有项目合并到一个IEnumerable<T>中(带有副本)。如其他答案中所述，使用Union而不是Concat删除重复项。|1782214|entityMap^0|4|X|0|0|3|5|J|E|1G|5|1O|6|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|N|$]]

All you need is this, for any <code>IEnumerable&lt;IEnumerable&lt;T&gt;&gt; lists</code> :

<pre><code>var combined = lists.Aggregate((l1, l2) =&gt; l1.Concat(l2));
</code></pre>

This will combine all the items in <code>lists</code> into one <code>IEnumerable&lt;T&gt;</code> (with duplicates). Use <code>Union</code> instead of <code>Concat</code> to remove duplicates, as noted in the other answers.

Say I have an arbitrary number of collections, each containing objects of the same type (for example, <code>List&lt;int&gt; foo</code> and <code>List&lt;int&gt; bar</code>). If these collections were themselves in a collection (e.g., of type <code>List&lt;List&lt;int&gt;&gt;</code>, I could use <code>SelectMany</code> to combine them all into one collection.

However, if these collections are not already in the same collection, it's my impression that I'd have to write a method like this:

<pre><code>public static IEnumerable&lt;T&gt; Combine&lt;T&gt;(params ICollection&lt;T&gt;[] toCombine)
{
 return toCombine.SelectMany(x =&gt; x);
}
</code></pre>

Which I'd then call like this:

<pre><code>var combined = Combine(foo, bar);
</code></pre>

Is there a clean, elegant way to combine (any number of) collections without having to write a utility method like <code>Combine</code> above? It seems simple enough that there should be a way to do it in LINQ, but perhaps not.

Elegant way to combine multiple collections of elements?

假设我有任意数量的集合，每个集合都包含相同类型的对象(例如，List<int> foo和List<int> bar)。如果这些集合本身在一个集合中(例如，类型为List<List<int>>，我可以使用SelectMany将它们组合到一个集合中)。但是，如果这些集合不在同一个集合中，我的印象是我必须编写一个这样的方法：public static IEnumerable<T> Combine<T>(

问组合多个元素集合的优雅方式？
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