blocks|key|1729097|text|我非常确定没有内置的方法可以做到这一点，但是通过链接Reverse和Skip可以很容易地做到这一点|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1729098|aCollection.Reverse().Skip(n).Reverse()|code-block|syntax|javascript|1729099|entityMap^0|Q|7|Y|4|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|$]]

I'm pretty sure there's no built-in method for this, but this can be done easily by chaining <code>Reverse</code> and <code>Skip</code>:

<pre><code>aCollection.Reverse().Skip(n).Reverse()
</code></pre>

blocks|key|1942592|text|我不相信这有一个内置的函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1942593|aCollection.Take(aCollection.Count+-+n)|offset|length|style|CODE|1942594|应该是合适的；取集合中的项目总数减去n应该跳过最后n个元素。|1942595|entityMap^0|0|0|13|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|K|$]]

I don't believe there's a built-in function for this.

<code>aCollection.Take(aCollection.Count - n)</code>

should be suitable; taking the total number of items in the collection minus n should skip the last n elements.

blocks|key|1729198|text|这将比使用双反向的解决方案效率高得多，因为它只创建一个列表，并且只枚举该列表一次。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1729199|public+static+class+Extensions
{
+++static+IEnumerable<T>+Leave<T>(this+IEnumerable<T>+items,+int+numToSkip)
+++{
++++++var+list+=+items.ToList();
++++++//+Assert+numToSkip+<=+list+count.
++++++list.RemoveRange(list.Count+-+numToSkip,+numToSkip);
++++++return+List
+++}
}


string+alphabet+=+"abcdefghijklmnopqrstuvwxyz";
var+chars+=+alphabet.Leave(10);+//+abcdefghijklmnop|code-block|syntax|javascript|1729200|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|$]]

This will be much more efficient than the solutions with a double-reverse, since it creates only one list and only enumerates the list once.

<pre><code>public static class Extensions
{
 static IEnumerable&lt;T&gt; Leave&lt;T&gt;(this IEnumerable&lt;T&gt; items, int numToSkip)
 {
 var list = items.ToList();
 // Assert numToSkip &lt;= list count.
 list.RemoveRange(list.Count - numToSkip, numToSkip);
 return List
 }
}


string alphabet = "abcdefghijklmnopqrstuvwxyz";
var chars = alphabet.Leave(10); // abcdefghijklmnop
</code></pre>

blocks|key|1729244|text|与IEnumerable哲学保持一致，对于没有实现ICollection的情况，只需枚举一次，您可以使用以下扩展方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1729245|public+static+IEnumerable<T>+Leave<T>(this+ICollection<T>+src,+int+drop)+=>+src.Take(src.Count+-+drop);

public+static+IEnumerable<T>+Leave<T>(this+IEnumerable<T>+src,+int+drop)+{
++++IEnumerable<T>+IEnumHelper()+{
++++++++using+(var+esrc+=+src.GetEnumerator())+{
++++++++++++var+buf+=+new+Queue<T>();
++++++++++++while+(drop--+>+0)
++++++++++++++++if+(esrc.MoveNext())
++++++++++++++++++++buf.Enqueue(esrc.Current);
++++++++++++++++else
++++++++++++++++++++break;

++++++++++++while+(esrc.MoveNext())+{
++++++++++++++++buf.Enqueue(esrc.Current);
++++++++++++++++yield+return+buf.Dequeue();
++++++++++++}
++++++++}
++++}

++++return+(src+is+ICollection<T>+csrc)+?+csrc.Leave(drop)+:+IEnumHelper();
}|code-block|syntax|javascript|1729246|entityMap^0|1|B|P|B|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|$]]

Keeping with the <code>IEnumerable</code> philosphy, and going through the enumeration once for cases where <code>ICollection</code> isn't implemented, you can use these extension methods:

<pre><code>public static IEnumerable&lt;T&gt; Leave&lt;T&gt;(this ICollection&lt;T&gt; src, int drop) =&gt; src.Take(src.Count - drop);

public static IEnumerable&lt;T&gt; Leave&lt;T&gt;(this IEnumerable&lt;T&gt; src, int drop) {
 IEnumerable&lt;T&gt; IEnumHelper() {
 using (var esrc = src.GetEnumerator()) {
 var buf = new Queue&lt;T&gt;();
 while (drop-- &gt; 0)
 if (esrc.MoveNext())
 buf.Enqueue(esrc.Current);
 else
 break;

 while (esrc.MoveNext()) {
 buf.Enqueue(esrc.Current);
 yield return buf.Dequeue();
 }
 }
 }

 return (src is ICollection&lt;T&gt; csrc) ? csrc.Leave(drop) : IEnumHelper();
}
</code></pre>

blocks|key|1729263|text|目前，C#定义了一个TakeLast(n)方法，该方法从字符串的末尾获取字符。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1729264|查看此处：https://msdn.microsoft.com/en-us/library/hh212114(v=vs.103).aspx|1729265|entityMap|0|LINK|mutability|MUTABLE|url|https://msdn.microsoft.com/en-us/library/hh212114(v=vs.103).aspx^0|A|B|0|5|1S|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@$9|T|A|U|1|V]]|E|$]]|$1|H|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

Currently, C# has a <code>TakeLast(n)</code> method defined which takes characters from the end of the string.

See here: <a href="https://msdn.microsoft.com/en-us/library/hh212114(v=vs.103).aspx" rel="nofollow noreferrer">https://msdn.microsoft.com/en-us/library/hh212114(v=vs.103).aspx</a>

Using Linq; how can I do the "opposite" of Take?

I.e. instead of getting the first n elements such as in

<pre><code>aCollection.Take(n)
</code></pre>

I want to get everything but the last n elements. Something like

<pre><code>aCollection.Leave(n)
</code></pre>

(Don't ask why :-)

Edit

I suppose I can do it this way <code>aCollection.TakeWhile((x, index) =&gt; index &lt; aCollection.Count - n)</code> Or in the form of an extension 

<pre><code>public static IEnumerable&lt;TSource&gt; Leave&lt;TSource&gt;(this IEnumerable&lt;TSource&gt; source, int n) 
{ 
 return source.TakeWhile((x, index) =&gt; index &lt; source.Count() - n); 
}
</code></pre>

But in the case of Linq to SQL or NHibernate Linq it would have been nice if the generated SQL took care of it and generated something like (for SQL Server/T-SQL) 

<code>SELECT TOP(SELECT COUNT(*) -@n FROM ATable) * FROM ATable</code> Or some other more clever SQL implementation.

I suppose there is nothing like it?
(But the edit was actually not part of the question.)

Linq: The "opposite" of Take?

SQL Server

使用Linq；我如何做Take的“对立面”？也就是说，不是获取前n个元素，例如aCollection.Take(n)我想要得到除最后n个元素之外的所有内容。就像这样aCollection.Leave(n)(不要问为什么:-)编辑我想我可以这样做，aCollection.TakeWhile((x, index) => i...

问Linq: Take的“对立面”？
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