blocks|key|344661|text|return+list.Where((x,+i)+=>+i+%25+nStep+==+0);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|344662|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>return list.Where((x, i) =&gt; i % nStep == 0);
</code></pre>

blocks|key|3334939|text|我知道这很“老派”，但为什么不直接使用带有loop+=+n的for循环呢？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3334940|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

I know it's "old school," but why not just use a for loop with stepping = n?

blocks|key|3122052|text|听起来像是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3122053|IEnumerator<T>+GetNth<T>(List<T>+list,+int+n)+{
++for+(int+i=0;+i<list.Count;+i%2B=n)
++++yield+return+list[i]
}|code-block|syntax|javascript|3122054|就能达到目的。我认为没有必要使用Linq或lambda表达式。|3122055|编辑：|offset|length|style|BOLD|3122056|搞定|3122057|public+static+class+MyListExtensions+{
++public+static+IEnumerable<T>+GetNth<T>(this+List<T>+list,+int+n)+{
++++for+(int+i=0;+i<list.Count;+i%2B=n)
++++++yield+return+list[i];
++}
}|3122058|你用一种LINQish的方式写|3122059|from+var+element+in+MyList.GetNth(10)+select+element;|3122060|第二次编辑|3122061|为了让它更具LINQish|3122062|from+var+i+in+Range(0,+((myList.Length-1)/n)%2B1)+select+list[n*i];|3122063|entityMap^0|0|0|0|0|3|0|0|0|0|0|0|5|0|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|16|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|17|8|@$K|18|L|19|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1A|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1B|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1C|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1D|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1E|8|@$K|1F|L|1G|M|N]]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1H|8|@]|9|@]|A|$]]|$1|10|3|11|5|D|7|1I|8|@]|9|@]|A|$E|F]]|$1|12|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]]|13|$]]

Sounds like

<pre><code>IEnumerator&lt;T&gt; GetNth&lt;T&gt;(List&lt;T&gt; list, int n) {
 for (int i=0; i&lt;list.Count; i+=n)
 yield return list[i]
}
</code></pre>

would do the trick. I do not see the need to use Linq or a lambda expressions.

EDIT:

Make it 

<pre><code>public static class MyListExtensions {
 public static IEnumerable&lt;T&gt; GetNth&lt;T&gt;(this List&lt;T&gt; list, int n) {
 for (int i=0; i&lt;list.Count; i+=n)
 yield return list[i];
 }
}
</code></pre>

and you write in a LINQish way

<pre><code>from var element in MyList.GetNth(10) select element;
</code></pre>

2nd Edit:

To make it even more LINQish

<pre><code>from var i in Range(0, ((myList.Length-1)/n)+1) select list[n*i];
</code></pre>

blocks|key|346041|text|您可以使用Where重载，它将索引与元素一起传递|type|unstyled|depth|inlineStyleRanges|entityRanges|data|346042|var+everyFourth+=+list.Where((x,i)+=>+i+%25+4+==+0);|code-block|syntax|javascript|346043|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 can use the Where overload which passes the index along with the element

<pre><code>var everyFourth = list.Where((x,i) =&gt; i % 4 == 0);
</code></pre>

blocks|key|349493|text|For循环|type|unstyled|depth|inlineStyleRanges|entityRanges|data|349494|for(int+i+=+0;+i+<+list.Count;+i+%2B=+n)
++++//Nth+Item..|code-block|syntax|javascript|349495|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|$]]

For Loop 

<pre><code>for(int i = 0; i &lt; list.Count; i += n)
 //Nth Item..
</code></pre>

blocks|key|3122148|text|我认为如果你提供了一个linq扩展，你应该能够在最不特定的接口上操作，比如在IEnumerable上。当然，如果您追求速度，特别是对于大N，您可能会为索引访问提供重载。后者消除了迭代大量不需要的数据的需要，并且将比Where子句快得多。提供这两个重载可以让编译器选择最合适的变体。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3122149|public+static+class+LinqExtensions
{
++++public+static+IEnumerable<T>+GetNth<T>(this+IEnumerable<T>+list,+int+n)
++++{
++++++++if+(n+<+0)
++++++++++++throw+new+ArgumentOutOfRangeException("n");
++++++++if+(n+>+0)
++++++++{
++++++++++++int+c+=+0;
++++++++++++foreach+(var+e+in+list)
++++++++++++{
++++++++++++++++if+(c+%25+n+==+0)
++++++++++++++++++++yield+return+e;
++++++++++++++++c%2B%2B;
++++++++++++}
++++++++}
++++}
++++public+static+IEnumerable<T>+GetNth<T>(this+IList<T>+list,+int+n)
++++{
++++++++if+(n+<+0)
++++++++++++throw+new+ArgumentOutOfRangeException("n");
++++++++if+(n+>+0)
++++++++++++for+(int+c+=+0;+c+<+list.Count;+c+%2B=+n)
++++++++++++++++yield+return+list[c];
++++}
}|code-block|syntax|javascript|3122150|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|$]]

I think if you provide a linq extension, you should be able to operate on the least specific interface, thus on IEnumerable. Of course, if you are up for speed especially for large N you might provide an overload for indexed access. The latter removes the need of iterating over large amounts of not needed data, and will be much faster than the Where clause. Providing both overloads lets the compiler select the most suitable variant.

<pre><code>public static class LinqExtensions
{
 public static IEnumerable&lt;T&gt; GetNth&lt;T&gt;(this IEnumerable&lt;T&gt; list, int n)
 {
 if (n &lt; 0)
 throw new ArgumentOutOfRangeException("n");
 if (n &gt; 0)
 {
 int c = 0;
 foreach (var e in list)
 {
 if (c % n == 0)
 yield return e;
 c++;
 }
 }
 }
 public static IEnumerable&lt;T&gt; GetNth&lt;T&gt;(this IList&lt;T&gt; list, int n)
 {
 if (n &lt; 0)
 throw new ArgumentOutOfRangeException("n");
 if (n &gt; 0)
 for (int c = 0; c &lt; list.Count; c += n)
 yield return list[c];
 }
}
</code></pre>

blocks|key|346052|text|我不确定是否可以使用LINQ表达式，但我知道您可以使用Where扩展方法来实现。例如，每隔5个项目获取一次：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|346053|List<T>+list+=+originalList.Where((t,i)+=>+(i+%25+5)+==+0).ToList();|code-block|syntax|javascript|346054|这将获得第一个项目和从那里开始的每五个项目。如果你想从第五项开始而不是第一项，你可以用4而不是0来比较。|346055|entityMap^0|R|5|0|0|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]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

I'm not sure if it's possible to do with a LINQ expression, but I know that you can use the <code>Where</code> extension method to do it. For example to get every fifth item:

<pre><code>List&lt;T&gt; list = originalList.Where((t,i) =&gt; (i % 5) == 0).ToList();
</code></pre>

This will get the first item and every fifth from there. If you want to start at the fifth item instead of the first, you compare with 4 instead of comparing with 0.

blocks|key|3122216|text|@belucha我喜欢这样，因为客户端代码可读性很好，编译器选择了最有效的实现。我将在此基础上减少对IReadOnlyList<T>的要求，并将部门节省为高性能的LINQ：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3122217|++++public+static+IEnumerable<T>+GetNth<T>(this+IEnumerable<T>+list,+int+n)+{
++++++++if+(n+<=+0)+throw+new+ArgumentOutOfRangeException(nameof(n),+n,+null);
++++++++int+i+=+n;
++++++++foreach+(var+e+in+list)+{
++++++++++++if+(%2B%2Bi+<+n)+{+//save+Division
++++++++++++++++continue;
++++++++++++}
++++++++++++i+=+0;
++++++++++++yield+return+e;
++++++++}
++++}

++++public+static+IEnumerable<T>+GetNth<T>(this+IReadOnlyList<T>+list,+int+n
++++++++,+int+offset+=+0)+{+//use+IReadOnlyList<T>
++++++++if+(n+<=+0)+throw+new+ArgumentOutOfRangeException(nameof(n),+n,+null);
++++++++for+(var+i+=+offset;+i+<+list.Count;+i+%2B=+n)+{
++++++++++++yield+return+list[i];
++++++++}
++++}|code-block|syntax|javascript|3122218|entityMap^0|1E|G|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

@belucha I like this, because the client code is very readable and the Compiler chooses the most efficient Implementation. I would build upon this by reducing the requirements to <code>IReadOnlyList&lt;T&gt;</code> and to save the Division for high-performance LINQ: 

<pre><code> public static IEnumerable&lt;T&gt; GetNth&lt;T&gt;(this IEnumerable&lt;T&gt; list, int n) {
 if (n &lt;= 0) throw new ArgumentOutOfRangeException(nameof(n), n, null);
 int i = n;
 foreach (var e in list) {
 if (++i &lt; n) { //save Division
 continue;
 }
 i = 0;
 yield return e;
 }
 }

 public static IEnumerable&lt;T&gt; GetNth&lt;T&gt;(this IReadOnlyList&lt;T&gt; list, int n
 , int offset = 0) { //use IReadOnlyList&lt;T&gt;
 if (n &lt;= 0) throw new ArgumentOutOfRangeException(nameof(n), n, null);
 for (var i = offset; i &lt; list.Count; i += n) {
 yield return list[i];
 }
 }
</code></pre>

blocks|key|3122155|text|private+static+readonly+string[]+sequence+=+"1,2,3,4,5,6,7,8,9,10,11,12,13,14,15".Split(',');

static+void+Main(string[]+args)
{
++++var+every4thElement+=+sequence
++++++.Where((p,+index)+=>+index+%25+4+==+0);

++++foreach+(string+p+in+every4thElement)
++++{
++++++++Console.WriteLine("{0}",+p);
++++}

++++Console.ReadKey();
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3122156|输出|unstyled|3122157|​|3122158|📷|atomic|offset|length|3122159|3122160|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/ce2291d615f3b27a192658b8dbf32f00.png|imageAlt^0|0|0|0|0|1|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|Z|8|@]|9|@]|A|$]]|$1|I|3|J|5|K|7|10|8|@]|9|@$L|11|M|12|1|13]]|A|$]]|$1|N|3|H|5|F|7|14|8|@]|9|@]|A|$]]|$1|O|3|-4|5|F|7|15|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V|W|-4]]]]

<pre><code>private static readonly string[] sequence = "1,2,3,4,5,6,7,8,9,10,11,12,13,14,15".Split(',');

static void Main(string[] args)
{
 var every4thElement = sequence
 .Where((p, index) =&gt; index % 4 == 0);

 foreach (string p in every4thElement)
 {
 Console.WriteLine("{0}", p);
 }

 Console.ReadKey();
}
</code></pre>

output

<a href="https://i.stack.imgur.com/6mURq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6mURq.png" alt="enter image description here"></a>

blocks|key|3335105|text|我想没有一个答案是正确的。所有的解决方案都从0开始。但是我想要有真正的第n个元素|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3335106|public+static+IEnumerable<T>+GetNth<T>(this+IList<T>+list,+int+n)
{
++++for+(int+i+=+n+-+1;+i+<+list.Count;+i+%2B=+n)
++++++++yield+return+list[i];
}|code-block|syntax|javascript|3335107|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|$]]

Imho no answer is right. All solutions begins from 0. But I want to have the real nth element

<pre><code>public static IEnumerable&lt;T&gt; GetNth&lt;T&gt;(this IList&lt;T&gt; list, int n)
{
 for (int i = n - 1; i &lt; list.Count; i += n)
 yield return list[i];
}
</code></pre>

I'm using .NET 3.5 and would like to be able to obtain every *<code>n</code>*th item from a List. I'm not bothered as to whether it's achieved using a lambda expression or LINQ.

Edit

Looks like this question provoked quite a lot of debate (which is a good thing, right?). The main thing I've learnt is that when you think you know every way to do something (even as simple as this), think again!

How can I get every nth item from a List<T>?

编译器

我使用的是.NET 3.5，希望能够从列表中获得每*n*个项目。我不关心它是使用lambda表达式还是LINQ实现的。编辑看起来这个问题引发了相当多的争论(这是一件好事，对吧？)我学到的主要事情是，当你认为你知道做某事的所有方法时(即使如此简单)，再想一想！

问如何从List<T>中获取每n个项目？
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