blocks|key|3102733|text|var+list+=+new+List<int>+{+3,+1,+0,+5+};
int+pos+=+list.IndexOf(list.Min());+//+returns+2|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3102734|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>var list = new List&lt;int&gt; { 3, 1, 0, 5 };
int pos = list.IndexOf(list.Min()); // returns 2
</code></pre>

blocks|key|3315642|text|由于您特别要求一个LINQ解决方案，而您得到的都是非LINQ解决方案，下面是一个LINQ解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3315643|List<int>+values+=+new+List<int>+{+3,+1,+0,+5+};

int+index+=
+++values
+++.Select((n,+i)+=>+new+{+Value+=+n,+Index+=+i+})
+++.OrderBy(n=>n.Value)
+++.First()
+++.Index;|code-block|syntax|javascript|3315644|然而，这并不意味着LINQ是这个问题的最佳解决方案……|3315645|编辑：|3315646|对于更复杂的代码，这会执行得更好：|3315647|int+index+=
+++values
+++.Select((n,+i)+=>+new+{+Value+=+n,+Index+=+i+})
+++.Aggregate((a,b)+=>+a.Value+<+b.Value+?+a+:+b)
+++.Index;|3315648|为了获得最佳性能，您可以使用一个简单的循环遍历这些项，同时跟踪最低的项：|3315649|int+index+=+0,+value+=+values[0];
for+(int+i+=+1;+i+<+values.Length;+i%2B%2B)+{
++if+(values[i]+<+value)+{
++++value+=+values[i];
++++index+=+i;
++}
}|3315650|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

As you specifically asked for a LINQ solution, and all you got was non-LINQ solutions, here's a LINQ solution:
<pre><code>List&lt;int&gt; values = new List&lt;int&gt; { 3, 1, 0, 5 };

int index =
 values
 .Select((n, i) =&gt; new { Value = n, Index = i })
 .OrderBy(n=&gt;n.Value)
 .First()
 .Index;
</code></pre>
That however doesn't mean that LINQ is the best solution for this problem...
<h3>Edit:</h3>
With a bit more complex code this performs a little better:
<pre><code>int index =
 values
 .Select((n, i) =&gt; new { Value = n, Index = i })
 .Aggregate((a,b) =&gt; a.Value &lt; b.Value ? a : b)
 .Index;
</code></pre>
To get the best performance, you would use a plain loop go get through the items, while you keep track of the lowest:
<pre><code>int index = 0, value = values[0];
for (int i = 1; i &lt; values.Length; i++) {
 if (values[i] &lt; value) {
 value = values[i];
 index = i;
 }
}
</code></pre>

blocks|key|3315728|text|获取位置的最佳方法是使用FindIndex。此函数仅适用于List<>|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3315729|示例|3315730|int+id+=+listMyObject.FindIndex(x+=>+x.Id+==+15);+|code-block|syntax|javascript|3315731|如果您有枚举器或数组，请使用此方法|3315732|int+id+=+myEnumerator.ToList().FindIndex(x+=>+x.Id+==+15);+|3315733|或|3315734|+++int+id+=+myArray.ToList().FindIndex(x+=>+x.Id+==+15);+|3315735|entityMap^0|C|9|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|13|8|@]|D|@]|E|$]]|$1|S|3|T|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|U|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|V|$]]

The best way to catch the position is by <code>FindIndex</code> 
This function is available only for List&lt;>

Example

<pre><code>int id = listMyObject.FindIndex(x =&gt; x.Id == 15); 
</code></pre>

If you have enumerator or array use this way

<pre><code>int id = myEnumerator.ToList().FindIndex(x =&gt; x.Id == 15); 
</code></pre>

or 

<pre><code> int id = myArray.ToList().FindIndex(x =&gt; x.Id == 15); 
</code></pre>

blocks|key|337491|text|我同意LINQ不是这个问题的最佳解决方案，但这里有另一个变体O(n)。它不排序，并且只遍历列表一次。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|337492|var+list+=+new+List<int>+{+3,+1,+0,+5+};
int+pos+=+Enumerable.Range(0,+list.Count)
++++.Aggregate((a,+b)+=>+(list[a]+<+list[b])+?+a+:+b);+//+returns+2|code-block|syntax|javascript|337493|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 agree that LINQ isn't the best solution for this problem, but here's another variation that is O(n). It doesn't sort and only traverses the list once.

<pre><code>var list = new List&lt;int&gt; { 3, 1, 0, 5 };
int pos = Enumerable.Range(0, list.Count)
 .Aggregate((a, b) =&gt; (list[a] &lt; list[b]) ? a : b); // returns 2
</code></pre>

blocks|key|3102750|text|var+data+=+new+List<int>+{+3,+1,+0,+5+};

var+result+=+Enumerable.Range(0,+data.Count).OrderBy(n+=>+data[n]).First();|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3102751|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>var data = new List&lt;int&gt; { 3, 1, 0, 5 };

var result = Enumerable.Range(0, data.Count).OrderBy(n =&gt; data[n]).First();
</code></pre>

blocks|key|3102914|text|一个列表可以包含多个等于最小值的元素(见下文)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3102915|我编写的泛型扩展方法.FindEveryIndex()可以处理整数、字符串……非常灵活，因为您可以将您的条件指定为Lambda表达式。|offset|length|style|CODE|3102916|另一个优点是，它返回匹配条件的所有索引的列表，而不仅仅是第一个元素。|3102917|关于您的问题：最小值可返回为：|3102918|var+lst+=+new+List<int>()+{+1,+2,+1,+3,+4,+1+};++//+example+list
var+minimum+=+lst.Min();++//+get+the+minumum+value+of+lst
var+idx+=+lst.FindEveryIndex(x+=>+x+==+minimum);++//+finds+all+indices+matching+condition
Console.WriteLine($"Output:+{String.Join(',',+idx.ToArray())}");++//+show+list+of+indices|code-block|syntax|javascript|3102919|它将返回索引0、2和5，因为lst1中的最小值为1|3102920|3102921|输出:+0,2,5|blockquote|3102922|3102923|3102924|示例2:|BOLD|3102925|void+Main()
{+++
++++//+working+with+list+of+integers
++++var+lst1+=+new+List<int>()+{+1,+2,+1,+3,+4,+1+};
++++lst1.FindEveryIndex(x+=>+x==1).Dump("Find+1");+++//+finds+indices:+[0,+2,+5]
++++lst1.FindEveryIndex(x+=>+x==2).Dump("Find+2");+++//+finds+index:+[1]
++++lst1.FindEveryIndex(x+=>+x==9).Dump("Find+9");+++//+returns+[-1]

++++//+working+with+list+of+strings
++++var+lst2+=+new+List<string>()+{+"A",+"B",+"A",+"C",+"D",+"A"};
++++lst2.FindEveryIndex(x+=>+x=="A").Dump("Find+A");+++//+finds+indices:+[0,+2,+5]
++++lst2.FindEveryIndex(x+=>+x=="B").Dump("Find+B");+++//+finds+index:+[1]
++++lst2.FindEveryIndex(x+=>+x=="X").Dump("Find+X");+++//+returns+[-1]
}|3102926|扩展类：|3102927|public+static+class+Extension
{
++++//+using+System.Collections.Generic;
++++public+static+IEnumerable<int>+FindEveryIndex<T>(this+IEnumerable<T>+items,+
+++++++++++++++++++++++++++++++++++++++++++++++++++++Predicate<T>+predicate)
++++{
++++++++int+index+=+0;+bool+found+=+false;
++++++++foreach+(var+item+in+items)
++++++++{
++++++++++++if+(predicate(item))
++++++++++++{
++++++++++++++++found+=+true;+yield+return+index;
++++++++++++};
++++++++++++index%2B%2B;
++++++++}
++++++++if+(!found)+yield+return+-1;
++++}
}|3102928|注意：将这两个代码片段复制到LinqPad+C#程序中，它会立即生效。|3102929|或者，使用DotNetFiddle在线运行它。|3102930|entityMap|0|LINK|mutability|MUTABLE|url|https://dotnetfiddle.net/TWq29j^0|0|A|H|0|0|0|0|E|4|O|1|0|0|0|0|0|0|4|0|0|0|4|0|0|0|3|0|5|C|0|0^^$0|@$1|2|3|4|5|6|7|1J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1K|8|@$D|1L|E|1M|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|1N|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|1O|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|1P|8|@]|9|@]|A|$O|P]]|$1|Q|3|R|5|6|7|1Q|8|@$D|1R|E|1S|F|G]|$D|1T|E|1U|F|G]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1V|8|@]|9|@]|A|$]]|$1|T|3|U|5|V|7|1W|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1X|8|@]|9|@]|A|$]]|$1|X|3|-4|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|6|7|1Z|8|@$D|20|E|21|F|10]]|9|@]|A|$]]|$1|11|3|12|5|N|7|22|8|@]|9|@]|A|$O|P]]|$1|13|3|14|5|6|7|23|8|@$D|24|E|25|F|10]]|9|@]|A|$]]|$1|15|3|16|5|N|7|26|8|@]|9|@]|A|$O|P]]|$1|17|3|18|5|6|7|27|8|@$D|28|E|29|F|10]]|9|@]|A|$]]|$1|19|3|1A|5|6|7|2A|8|@]|9|@$D|2B|E|2C|1|2D]]|A|$]]|$1|1B|3|-4|5|6|7|2E|8|@]|9|@]|A|$]]]|1C|$1D|$5|1E|1F|1G|A|$1H|1I]]]]

A list can contain multiple elements which are equal to the minimum value (see below).
The generic extension method <code>.FindEveryIndex()</code> I wrote works with integers, strings, ... and is quite flexible because you can specify your condition as Lambda expression.
Another advantage is that it returns a list of all indices matching the condition, not just the first element.
Regarding your question: The minimum can be returned as:
<pre><code>var lst = new List&lt;int&gt;() { 1, 2, 1, 3, 4, 1 }; // example list
var minimum = lst.Min(); // get the minumum value of lst
var idx = lst.FindEveryIndex(x =&gt; x == minimum); // finds all indices matching condition
Console.WriteLine($&quot;Output: {String.Join(',', idx.ToArray())}&quot;); // show list of indices
</code></pre>
It will return the indices 0, 2 and 5, because the minimum in <code>lst1</code> is <code>1</code>:
<blockquote>
Output: 0,2,5
</blockquote>
Example 2:
<pre><code>void Main()
{ 
 // working with list of integers
 var lst1 = new List&lt;int&gt;() { 1, 2, 1, 3, 4, 1 };
 lst1.FindEveryIndex(x =&gt; x==1).Dump(&quot;Find 1&quot;); // finds indices: [0, 2, 5]
 lst1.FindEveryIndex(x =&gt; x==2).Dump(&quot;Find 2&quot;); // finds index: [1]
 lst1.FindEveryIndex(x =&gt; x==9).Dump(&quot;Find 9&quot;); // returns [-1]

 // working with list of strings
 var lst2 = new List&lt;string&gt;() { &quot;A&quot;, &quot;B&quot;, &quot;A&quot;, &quot;C&quot;, &quot;D&quot;, &quot;A&quot;};
 lst2.FindEveryIndex(x =&gt; x==&quot;A&quot;).Dump(&quot;Find A&quot;); // finds indices: [0, 2, 5]
 lst2.FindEveryIndex(x =&gt; x==&quot;B&quot;).Dump(&quot;Find B&quot;); // finds index: [1]
 lst2.FindEveryIndex(x =&gt; x==&quot;X&quot;).Dump(&quot;Find X&quot;); // returns [-1]
}
</code></pre>
Extension class:
<pre><code>public static class Extension
{
 // using System.Collections.Generic;
 public static IEnumerable&lt;int&gt; FindEveryIndex&lt;T&gt;(this IEnumerable&lt;T&gt; items, 
 Predicate&lt;T&gt; predicate)
 {
 int index = 0; bool found = false;
 foreach (var item in items)
 {
 if (predicate(item))
 {
 found = true; yield return index;
 };
 index++;
 }
 if (!found) yield return -1;
 }
}
</code></pre>
Note: Copy the two code snippets into a LinqPad C# program and it works instantly.
Or, run it online with <a href="https://dotnetfiddle.net/TWq29j" rel="nofollow noreferrer">DotNetFiddle</a>.

blocks|key|3315595|text|List<int>+data+=+new+List<int>();
data.AddRange(new[]+{+3,+1,+0,+5+});
Console.WriteLine(data.IndexOf(data.Min()));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3315596|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>List&lt;int&gt; data = new List&lt;int&gt;();
data.AddRange(new[] { 3, 1, 0, 5 });
Console.WriteLine(data.IndexOf(data.Min()));
</code></pre>

blocks|key|3315652|text|int+min+=+0;
bool+minIsSet+=+false;

var+result+=+ints
++.Select(+(x,+i)+=>+new+{x,+i}
++.OrderBy(z+=>+z.x)
++.Select(z+=>+
++{
++++if+(!minIsSet)
++++{
++++++min+=+z.x;
++++++minIsSet+=+true;
++++}
++++return+z;
++}
++.TakeWhile(z+=>+z.x+==+min)
++.Select(z+=>+z.i);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3315653|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>int min = 0;
bool minIsSet = false;

var result = ints
 .Select( (x, i) =&gt; new {x, i}
 .OrderBy(z =&gt; z.x)
 .Select(z =&gt; 
 {
 if (!minIsSet)
 {
 min = z.x;
 minIsSet = true;
 }
 return z;
 }
 .TakeWhile(z =&gt; z.x == min)
 .Select(z =&gt; z.i);
</code></pre>

blocks|key|339919|text|我不一定推荐这种CPS风格的代码，但与使用OrderBy的解决方案不同，它可以工作并且是O(n)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|339920|var+minIndex+=+list.Aggregate(
++++new+{+i+=+0,+mini+=+-1,+minv+=+int.MaxValue+},
++++(min,+x)+=>+(min.minv+>+x)
++++++++?+new+{+i+=+min.i+%2B+1,+mini+=+min.i,+minv+=+x+}
++++++++:+new+{+i+=+min.i+%2B+1,+mini+=+min.mini,+minv+=+min.minv+})
++++.mini;|code-block|syntax|javascript|339921|如果您想要最后一个最小副本，而不是第一个副本，请将>更改为>=。|339922|使用.minv获取最小值，或者都不使用获取包含索引和最小值的二元组。|339923|我迫不及待地想让.NET在4.0中获得元组。|339924|entityMap^0|0|0|0|0|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I don't necessarily recommend this CPS-style code, but it works and is O(n), unlike the solutions that use OrderBy:

<pre><code>var minIndex = list.Aggregate(
 new { i = 0, mini = -1, minv = int.MaxValue },
 (min, x) =&gt; (min.minv &gt; x)
 ? new { i = min.i + 1, mini = min.i, minv = x }
 : new { i = min.i + 1, mini = min.mini, minv = min.minv })
 .mini;
</code></pre>

Change > to >= if you want the last minimum duplicate, not the first.

Use .minv to get the minimum value or neither to get a 2-tuple with both the index and the minimum value.

I can't wait for .NET to get tuples in 4.0.

blocks|key|3315620|text|List<int>.Enumerator+e+=+l.GetEnumerator();
int+p+=+0,+min+=+int.MaxValue,+pos+=+-1;
while+(e.MoveNext())
{
++++if+(e.Current+<+min)
++++{
++++++++min+=+e.Current;
++++++++pos+=+p;
++++}
++++%2B%2Bp;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3315621|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>List&lt;int&gt;.Enumerator e = l.GetEnumerator();
int p = 0, min = int.MaxValue, pos = -1;
while (e.MoveNext())
{
 if (e.Current &lt; min)
 {
 min = e.Current;
 pos = p;
 }
 ++p;
}
</code></pre>

I have a List with numbers,
and I'd like to find the position of the minimum (not value) using LINQ
Example:
<pre><code>var lst = new List&lt;int&gt;() { 3, 1, 0, 5 };
</code></pre>
Now I am looking for a function returning me
<blockquote>
output = 2
</blockquote>
because the minimum is at position 2 in the list.

Get List<> element position in c# using LINQ

我有一个包含数字的列表，我想使用LINQ找到最小值(而不是值)的位置示例：var lst = new List<int>() { 3, 1, 0, 5 };现在我正在寻找一个返回给我的函数输出=2因为最小值在列表中的位置2。

问使用LINQ获取c#中的List<>元素位置
EN

回答 10

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