如何对集合中所有对象的属性执行.Max()并返回具有最大值的对象 [英] How to perform .Max() on a property of all objects in a collection and return the object with maximum value

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问题描述

我有一个具有两个int属性的对象列表.该列表是另一个linq查询的输出.对象:

I have a list of objects that have two int properties. The list is the output of another linq query. The object:

public class DimensionPair  
{
    public int Height { get; set; }
    public int Width { get; set; }
}

我想在列表中查找并返回具有最大Height属性值的对象.

I want to find and return the object in the list which has the largest Height property value.

我可以设法获得Height值的最大值,而不是对象本身.

I can manage to get the highest value of the Height value but not the object itself.

我可以用Linq做到这一点吗?怎么样?

Can I do this with Linq? How?

推荐答案

我们有一个 MoreLINQ 中完全做到这一点.您可以在那儿查看实现,但是基本上是遍历数据的情况,记住我们到目前为止所看到的最大元素以及在投影下产生的最大值.

We have an extension method to do exactly this in MoreLINQ. You can look at the implementation there, but basically it's a case of iterating through the data, remembering the maximum element we've seen so far and the maximum value it produced under the projection.

对于您而言,您可以执行以下操作:

In your case you'd do something like:

var item = items.MaxBy(x => x.Height);

与Mehrdad的第二个解决方案(与MaxBy基本上相同)相比,这比此处提供的任何解决方案都要好(IMO):

This is better (IMO) than any of the solutions presented here other than Mehrdad's second solution (which is basically the same as MaxBy):

  • 它是O(n),与先前接受的答案不同,后者在每次迭代中都找到最大值(使其成为O( n ^ 2))
  • 订购解决方案为O(n log n)
  • Max值,然后找到具有该值的第一个元素为O(n),但对该序列进行两次迭代.在可能的情况下,您应该以单遍方式使用LINQ.
  • 与聚合版本相比,它的阅读和理解要简单得多,并且每个元素仅评估一次投影
  • It's O(n) unlike the previous accepted answer which finds the maximum value on every iteration (making it O(n^2))
  • The ordering solution is O(n log n)
  • Taking the Max value and then finding the first element with that value is O(n), but iterates over the sequence twice. Where possible, you should use LINQ in a single-pass fashion.
  • It's a lot simpler to read and understand than the aggregate version, and only evaluates the projection once per element

这篇关于如何对集合中所有对象的属性执行.Max()并返回具有最大值的对象的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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