blocks|key|1303619|text|int+minAccountLevel+=+int.MaxValue;
int+maxAccountLevel+=+int.MinValue;
foreach+(DataRow+dr+in+table.Rows)
{
++++int+accountLevel+=+dr.Field<int>("AccountLevel");
++++minAccountLevel+=+Math.Min(minAccountLevel,+accountLevel);
++++maxAccountLevel+=+Math.Max(maxAccountLevel,+accountLevel);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1303620|是的，这确实是最快的方法。使用Linq+Min和Max扩展总是比较慢，因为您必须迭代两次。您可能会使用Linq+Aggregate，但是其语法不会比现在更漂亮。|unstyled|offset|length|style|CODE|1303621|entityMap^0|0|K|3|O|3|1K|9|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]|$G|Q|H|R|I|J]|$G|S|H|T|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|U|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>int minAccountLevel = int.MaxValue;
int maxAccountLevel = int.MinValue;
foreach (DataRow dr in table.Rows)
{
 int accountLevel = dr.Field&lt;int&gt;("AccountLevel");
 minAccountLevel = Math.Min(minAccountLevel, accountLevel);
 maxAccountLevel = Math.Max(maxAccountLevel, accountLevel);
}
</code></pre>

Yes, this really is the fastest way. Using the Linq <code>Min</code> and <code>Max</code> extensions will always be slower because you have to iterate twice. You could potentially use Linq <code>Aggregate</code>, but the syntax isn't going to be much prettier than this already is.

blocks|key|1088531|text|在datatable上的Easiar方法可以是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1088532|int+minLavel+=+Convert.ToInt32(dt.Compute("min([AccountLevel])",+string.Empty));|code-block|syntax|javascript|1088533|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|$]]

Easiar approach on datatable could be:

<pre><code>int minLavel = Convert.ToInt32(dt.Compute("min([AccountLevel])", string.Empty));
</code></pre>

blocks|key|1088387|text|使用LINQ。只要将行集合转换为IEnumerable，它在数据表上就可以很好地工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1088388|List<int>+levels+=+AccountTable.AsEnumerable().Select(al+=>+al.Field<int>("AccountLevel")).Distinct().ToList();
int+min+=+levels.Min();
int+max+=+levels.Max();|code-block|syntax|javascript|1088389|编辑以修复语法；在DataTables上使用LINQ时很棘手，聚合函数也很有趣。|1088390|是的，这可以通过一个查询完成，但是您需要生成一个结果列表，然后在单独的语句中使用.Min()和.Max()作为聚合函数。|1088391|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Use LINQ. It works just fine on datatables, as long as you convert the rows collection to an IEnumerable.

<pre><code>List&lt;int&gt; levels = AccountTable.AsEnumerable().Select(al =&gt; al.Field&lt;int&gt;("AccountLevel")).Distinct().ToList();
int min = levels.Min();
int max = levels.Max();
</code></pre>

Edited to fix syntax; it's tricky when using LINQ on DataTables, and aggregating functions are fun, too.

Yes, it can be done with one query, but you will need to generate a list of results, then use .Min() and .Max() as aggregating functions in separate statements.

blocks|key|1088685|text|这对我来说很好。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1088686|int++max+=+Convert.ToInt32(datatable_name.AsEnumerable()
++++++++++++++++++++++++.Max(row+=>+row["column_Name"]));|code-block|syntax|javascript|1088687|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 worked fine for me

<pre><code>int max = Convert.ToInt32(datatable_name.AsEnumerable()
 .Max(row =&gt; row["column_Name"]));
</code></pre>

blocks|key|1088323|text|最有效的方法(信不信由你)是创建两个变量并编写一个for循环。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1088324|entityMap^0|P|3|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

The most efficient way to do this (believe it or not) is to make two variables and write a <code>for</code> loop.

blocks|key|1088500|text|var+answer+=+accountTable.Aggregate(new+{+Min+=+int.MinValue,+Max+=+int.MaxValue+},+
++++++++++++++++++++++++++++++++++++++++(a,+b)+=>+new+{+Min+=+Math.Min(a.Min,+b.Field<int>("AccountLevel")),
++++++++++++++++++++++++++++++++++++++++++++++++++++++++Max+=+Math.Max(a.Max,+b.Field<int>("AccountLevel"))+});
int+min+=+answer.Min;
int+max+=+answer.Max;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1088501|1次迭代，linq风格:)|unstyled|1088502|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>var answer = accountTable.Aggregate(new { Min = int.MinValue, Max = int.MaxValue }, 
 (a, b) =&gt; new { Min = Math.Min(a.Min, b.Field&lt;int&gt;("AccountLevel")),
 Max = Math.Max(a.Max, b.Field&lt;int&gt;("AccountLevel")) });
int min = answer.Min;
int max = answer.Max;
</code></pre>

1 iteration, linq style :)

blocks|key|1303833|text|完成此操作的另一种方法是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1303834|int+minLavel+=+Convert.ToInt32(dt.Select("AccountLevel=min(AccountLevel)")[0][0]);|code-block|syntax|javascript|1303835|我不确定性能部分，但这确实给出了正确的输出|1303836|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

another way of doing this is 

<pre><code>int minLavel = Convert.ToInt32(dt.Select("AccountLevel=min(AccountLevel)")[0][0]);
</code></pre>

I am not sure on the performace part but this does give the correct output

blocks|key|1088699|text|var+min+=+dt.AsEnumerable().Min(row+=>+row["AccountLevel"]);
var+max+=+dt.AsEnumerable().Max(row+=>+row["AccountLevel"]);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1088700|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 min = dt.AsEnumerable().Min(row =&gt; row["AccountLevel"]);
var max = dt.AsEnumerable().Max(row =&gt; row["AccountLevel"]);
</code></pre>

blocks|key|1303947|text|Session["MinDate"]+=+dtRecord.Compute("Min(AccountLevel)",+string.Empty);
Session["MaxDate"]+=+dtRecord.Compute("Max(AccountLevel)",+string.Empty);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1303948|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>Session["MinDate"] = dtRecord.Compute("Min(AccountLevel)", string.Empty);
Session["MaxDate"] = dtRecord.Compute("Max(AccountLevel)", string.Empty);
</code></pre>

blocks|key|1303935|text|在性能方面，这应该是可比较的。使用Select语句和Sort来获取列表，然后选择第一个或最后一个(取决于您的排序顺序)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1303936|var+col+=+dt.Select("AccountLevel",+"AccountLevel+ASC");

var+min+=+col.First();
var+max+=+col.Last();|code-block|syntax|javascript|1303937|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|$]]

Performance wise, this should be comparable. Use Select statement and Sort to get a list and then pick the first or last (depending on your sort order).

<pre><code>var col = dt.Select("AccountLevel", "AccountLevel ASC");

var min = col.First();
var max = col.Last();
</code></pre>

blocks|key|1303786|text|我不知道我的解决方案如何将性能智慧与以前的答案进行比较。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1303787|我知道最初的问题是:在DataTable对象中获取最小和最大值的最快方法是什么，这可能是一种方法：|1303788|DataView+view+=+table.DefaultView;
view.Sort+=+"AccountLevel";
DataTable+sortedTable+=+view.ToTable();
int+min+=+sortedTable.Rows[0].Field<int>("AccountLevel");
int+max+=+sortedTable.Rows[sortedTable.Rows.Count-1].Field<int>("AccountLevel");|code-block|syntax|javascript|1303789|这是一种简单的方法，可以在不循环的情况下实现相同的结果。但性能需要与之前的答案进行比较。我想我最喜欢赛昂猫的答案。|1303790|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I don't know how my solution compares performance wise to previous answers.

I understand that the initial question was: What is the fastest way to get min and max values in a DataTable object, this may be one way of doing it:

<pre><code>DataView view = table.DefaultView;
view.Sort = "AccountLevel";
DataTable sortedTable = view.ToTable();
int min = sortedTable.Rows[0].Field&lt;int&gt;("AccountLevel");
int max = sortedTable.Rows[sortedTable.Rows.Count-1].Field&lt;int&gt;("AccountLevel");
</code></pre>

It's an easy way of achieving the same result without looping. But performance will need to be compared with previous answers. Thought I love Cylon Cats answer most.

For the following datatable column, what is the fastest way to get the min and max values?

<pre><code>AccountLevel 
0 
1 
2 
3 
</code></pre>

How to select min and max values of a column in a datatable?

对于下面的datatable列，获取最小和最大值的最快方法是什么？AccountLevel  0  1  2  3 

问如何在数据表中选择列的最小和最大值？
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