blocks|key|1742982|text|foreach+(var+v+in+structure)+
{+++++
++++var+group+=+groups.Single(g+=>+g.Key+==+v.+???+);
++++v.ListOfSmth+=+group.ToList();
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1742983|unstyled|1742984|首先，您需要选择所需的组。然后，您可以在组上使用的ToList方法。IGrouping是这些值的IEnumerable。|offset|length|style|CODE|1742985|entityMap^0|0|0|P|6|Y|9|1C|B|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|E|7|P|8|@$H|Q|I|R|J|K]|$H|S|I|T|J|K]|$H|U|I|V|J|K]]|9|@]|A|$]]|$1|L|3|-4|5|E|7|W|8|@]|9|@]|A|$]]]|M|$]]

<pre><code>foreach (var v in structure) 
{ 
 var group = groups.Single(g =&gt; g.Key == v. ??? );
 v.ListOfSmth = group.ToList();
}
</code></pre>

<hr>

First you need to select the desired group. Then you can use the <code>ToList</code> method of on the group. The <code>IGrouping</code> is a <code>IEnumerable</code> of the values.

blocks|key|1956420|text|由于IGrouping<TKey,+TElement>实现了IEnumerable<TElement>，因此您可以使用SelectMany将所有IEnumerables放回一个IEnumerable中：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1956421|List<smth>+list+=+new+List<smth>();
IEnumerable<IGrouping<int,+smth>>+groups+=+list.GroupBy(x+=>+x.id);
IEnumerable<smth>+smths+=+groups.SelectMany(group+=>+group);
List<smth>+newList+=+smths.ToList();|code-block|syntax|javascript|1956422|下面是一个构建/运行的示例：https://dotnetfiddle.net/DyuaaP|1956423|此解决方案的视频评论：https://youtu.be/6BsU1n1KTdo|1956424|entityMap|0|LINK|mutability|MUTABLE|url|https://dotnetfiddle.net/DyuaaP|1|https://youtu.be/6BsU1n1KTdo^0|2|P|U|L|1N|A|20|C|2G|B|0|0|E|V|0|0|B|S|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1A|8|@]|D|@$9|1B|A|1C|1|1D]]|E|$]]|$1|M|3|N|5|6|7|1E|8|@]|D|@$9|1F|A|1G|1|1H]]|E|$]]|$1|O|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]]]

Since <code>IGrouping&lt;TKey, TElement&gt;</code> implements <code>IEnumerable&lt;TElement&gt;</code>, you can use <code>SelectMany</code> to put all the <code>IEnumerables</code> back into one <code>IEnumerable</code> all together:
<pre><code>List&lt;smth&gt; list = new List&lt;smth&gt;();
IEnumerable&lt;IGrouping&lt;int, smth&gt;&gt; groups = list.GroupBy(x =&gt; x.id);
IEnumerable&lt;smth&gt; smths = groups.SelectMany(group =&gt; group);
List&lt;smth&gt; newList = smths.ToList();
</code></pre>
Here's an example that builds/runs: <a href="https://dotnetfiddle.net/DyuaaP" rel="noreferrer">https://dotnetfiddle.net/DyuaaP</a>
Video commentary of this solution: <a href="https://youtu.be/6BsU1n1KTdo" rel="noreferrer">https://youtu.be/6BsU1n1KTdo</a>

blocks|key|1743172|text|来自IGrouping的定义：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1743173|IGrouping<out+TKey,+out+TElement>+:+IEnumerable<TElement>,+IEnumerable|code-block|syntax|javascript|1743174|你可以像这样遍历元素：|1743175|IEnumerable<IGrouping<int,+smth>>+groups+=+list.GroupBy(x+=>+x.ID)
foreach(IEnumerable<smth>+element+in+groups)
{
//do+something
}|1743176|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

From definition of IGrouping : 

<pre><code>IGrouping&lt;out TKey, out TElement&gt; : IEnumerable&lt;TElement&gt;, IEnumerable
</code></pre>

you can just iterate through elements like this:

<pre><code>IEnumerable&lt;IGrouping&lt;int, smth&gt;&gt; groups = list.GroupBy(x =&gt; x.ID)
foreach(IEnumerable&lt;smth&gt; element in groups)
{
//do something
}
</code></pre>

blocks|key|1743210|text|上述答案的更清晰版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1743211|IEnumerable<IGrouping<int,+ClassA>>+groups+=+list.GroupBy(x+=>+x.PropertyIntOfClassA);

foreach+(var+groupingByClassA+in+groups)
{
++++int+propertyIntOfClassA+=+groupingByClassA.Key;

++++//iterating+through+values
++++foreach+(var+classA+in+groupingByClassA)
++++{
++++++++int+key+=+classA.PropertyIntOfClassA;
++++}
}|code-block|syntax|javascript|1743212|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|$]]

More clarified version of above answers:

<pre><code>IEnumerable&lt;IGrouping&lt;int, ClassA&gt;&gt; groups = list.GroupBy(x =&gt; x.PropertyIntOfClassA);

foreach (var groupingByClassA in groups)
{
 int propertyIntOfClassA = groupingByClassA.Key;

 //iterating through values
 foreach (var classA in groupingByClassA)
 {
 int key = classA.PropertyIntOfClassA;
 }
}
</code></pre>

blocks|key|1743318|text|假设您有如下的MyPayments类|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1743319|+public+class+Mypayment
{
++++public+int+year+{+get;+set;+}
++++public+string+month+{+get;+set;+}
++++public+string+price+{+get;+set;+}
++++public+bool+ispaid+{+get;+set;+}
}|code-block|syntax|javascript|1743320|你有一个MyPayments的列表|1743321|public+List<Mypayment>+mypayments+{+get;+set;+}|1743322|并且您希望按年份对列表进行分组。您可以像这样使用linq：|1743323|List<List<Mypayment>>+mypayments+=+(from+IGrouping<int,+Mypayment>+item+in+yearGroup
++++++++++++++++++++++++++++++++++++++++++++++++let+mypayments1+=+(from+_payment+in+UserProjects.mypayments
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++where+_payment.year+==+item.Key
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++select+_payment).ToList()
++++++++++++++++++++++++++++++++++++++++++++++++select+mypayments1).ToList();|1743324|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Assume that you have MyPayments class like
<pre><code> public class Mypayment
{
 public int year { get; set; }
 public string month { get; set; }
 public string price { get; set; }
 public bool ispaid { get; set; }
}
</code></pre>
and you have a list of MyPayments
<pre><code>public List&lt;Mypayment&gt; mypayments { get; set; }
</code></pre>
and you want group the list by year. You can use linq like this:
<pre><code>List&lt;List&lt;Mypayment&gt;&gt; mypayments = (from IGrouping&lt;int, Mypayment&gt; item in yearGroup
 let mypayments1 = (from _payment in UserProjects.mypayments
 where _payment.year == item.Key
 select _payment).ToList()
 select mypayments1).ToList();
</code></pre>

blocks|key|1956693|text|只需执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1956694|//+this+will+"split"+the+list+into+groups
var+groups+=+list.GroupBy(x+=>+x.ID);

//+groups+is+a+"collection+of+lists"
foreach+(var+sublist+in+groups)
{
++//+now+the+sublist+is+only+a+part+of+the+original+list
++//+to+get+which+is+the+value+of+ID,+you+can+use+sublist.Key
}|code-block|syntax|javascript|1956695|你不需要Select().GroupBy(expr)来创建“列表列表”，某种意义上。|offset|length|style|CODE|1956696|entityMap^0|0|0|4|M|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|$]]

Simply do this:
<pre><code>// this will &quot;split&quot; the list into groups
var groups = list.GroupBy(x =&gt; x.ID);

// groups is a &quot;collection of lists&quot;
foreach (var sublist in groups)
{
 // now the sublist is only a part of the original list
 // to get which is the value of ID, you can use sublist.Key
}
</code></pre>
You don't need <code>Select().GroupBy(expr)</code> makes &quot;a list of lists&quot;, kind of.

I have a question about <code>IGrouping</code> and the <code>Select()</code> method.

Let's say I've got an <code>IEnumerable&lt;IGrouping&lt;int, smth&gt;&gt;</code> in this way:

<pre><code>var groups = list.GroupBy(x =&gt; x.ID);
</code></pre>

where <code>list</code> is a <code>List&lt;smth&gt;</code>.

And now I need to pass values of each <code>IGrouping</code> to another list in some way:

<pre><code>foreach (var v in structure)
{
 v.ListOfSmth = groups.Select(...); // &lt;- ???
}
</code></pre>

<hr>

Can anybody suggest how to get the values (<code>List&lt;smth&gt;</code>) from an <code>IGrouping&lt;int, smth&gt;</code> in such a context?

How to get values from IGrouping

我有一个关于IGrouping和Select()方法的问题。假设我有一个这样的IEnumerable<IGrouping<int, smth>>：var groups = list.GroupBy(x => x.ID);其中list是一个List<smth>。现在我需要以某种方式将每个IGrouping的值传递给另一个...

问如何从IGrouping获取值
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