本文是小编为大家收集整理的关于使用LINQ获取列表的最高n％的处理/解决方法，可以参考本文帮助大家快速定位并解决问题，中文翻译不准确的可切换到English标签页查看源文。
问题描述
我一直在尝试找到一种方法，让Linq能够选择给定列表的顶部N％.我获得的最接近的是拍摄声明，其作品与最高百分比SQL语句类似，但不支持百分比.我敢肯定我错过了一些明显的东西，但我似乎看不到它.
推荐答案
假设源是ICollection<T>(而不仅仅是IEnumerable<T>)，您可以做这样的事情:
public static IEnumerable<T> TakePercent<T>(this ICollection<T> source, double percent)
{
int count = (int)(source.Count * percent / 100);
return source.Take(count);
}
请注意，它可以与IEnumerable<T>一起使用(使用Count()方法)，但是它将列举两次序列，通常被认为是一件坏事.
问题描述
I have been trying to find a way for LINQ to be able to select the top n% of a given list. The closest I have been able to get is the take statement which works similarly to the TOP PERCENT SQL statement but doesn't support percentages. I'm sure I'm missing something obvious but I just can't quite seem to see it.
推荐答案
Assuming the source is an ICollection<T> (and not just an IEnumerable<T>), you can do something like that:
public static IEnumerable<T> TakePercent<T>(this ICollection<T> source, double percent)
{
int count = (int)(source.Count * percent / 100);
return source.Take(count);
}
Note that it could work with an IEnumerable<T> (using the Count() method), but it would enumerate the sequence twice, which is usually considered a bad thing.
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