LINQ中的标准偏差

Question

LINQ是否为聚合SQL函数STDDEV() (标准差)建模？

如果没有,那么计算它的最简单/最佳实践方法是什么？

例:

  SELECT test_id, AVERAGE(result) avg, STDDEV(result) std 
    FROM tests
GROUP BY test_id

Answer 1

您可以自己进行扩展计算

public static class Extensions
{
    public static double StdDev(this IEnumerable<double> values)
    {
       double ret = 0;
       int count = values.Count();
       if (count  > 1)
       {
          //Compute the Average
          double avg = values.Average();

          //Perform the Sum of (value-avg)^2
          double sum = values.Sum(d => (d - avg) * (d - avg));

          //Put it all together
          ret = Math.Sqrt(sum / count);
       }
       return ret;
    }
}

如果您有一个人口样本而不是整个人口,那么您应该使用ret = Math.Sqrt(sum / (count - 1));.

转换为Chris Bennett从添加标准偏差到LINQ的扩展.

Answer 2

Dynami的答案有效,但会通过数据进行多次传递以获得结果.这是一种计算样本标准差的单程方法:

public static double StdDev(this IEnumerable<double> values)
{
    // ref: http://warrenseen.com/blog/2006/03/13/how-to-calculate-standard-deviation/
    double mean = 0.0;
    double sum = 0.0;
    double stdDev = 0.0;
    int n = 0;
    foreach (double val in values)
    {
        n++;
        double delta = val - mean;
        mean += delta / n;
        sum += delta * (val - mean);
    }
    if (1 < n)
        stdDev = Math.Sqrt(sum / (n - 1));

    return stdDev;
}

这是样本标准差,因为它除以n - 1.对于正常的标准偏差,您需要除以n.

这使用Welford的方法,与Average(x^2)-Average(x)^2方法相比具有更高的数值精度.

Answer 3

这将David Clarke的答案转换为与其他聚合LINQ函数(如Average)相同形式的扩展.

用法是: var stdev = data.StdDev(o => o.number)

public static class Extensions
{
    public static double StdDev<T>(this IEnumerable<T> list, Func<T, double> values)
    {
        // ref: https://stackoverflow.com/questions/2253874/linq-equivalent-for-standard-deviation
        // ref: http://warrenseen.com/blog/2006/03/13/how-to-calculate-standard-deviation/ 
        var mean = 0.0;
        var sum = 0.0;
        var stdDev = 0.0;
        var n = 0;
        foreach (var value in list.Select(values))
        {
            n++;
            var delta = value - mean;
            mean += delta / n;
            sum += delta * (value - mean);
        }
        if (1 < n)
            stdDev = Math.Sqrt(sum / (n - 1));

        return stdDev; 

    }
} 

Answer 4

var stddev = Math.Sqrt(data.Average(z=>z*z)-Math.Pow(data.Average(),2));