使用LINQ获得列表的前N个百分比 | 使用LINQ获得列表的前N个百分比

本文介绍了使用LINQ获得列表的前N个百分比的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我一直在努力寻找一种方法，使LINQ能够选择给定列表的前n％个.我能够获得的最接近的是take语句，该语句的用法类似于TOP PERCENT SQL语句，但不支持百分比.我确定我缺少明显的东西，但似乎看不到.

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

假设源是ICollection<T>(而不仅仅是IEnumerable<T>)，则可以执行以下操作:

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);
}

请注意，它可以与IEnumerable<T>一起使用(使用Count()方法)，但是会枚举两次序列，通常认为这是一件坏事.

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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