Standard Deviation in LINQ

Question

Does LINQ model the aggregate SQL function STDDEV() (standard deviation)?

If not, what is the simplest / best-practices way to calculate it?

Example:

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

Answer 1

You can make your own extension calculating it

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;
    }
}

If you have a sample of the population rather than the whole population, then you should use ret = Math.Sqrt(sum / (count - 1));.

Transformed into extension from Adding Standard Deviation to LINQ by Chris Bennett.

Answer 2

Dynami's answer works but makes multiple passes through the data to get a result. This is a single pass method that calculates the sample standard deviation:

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;
}

This is the sample standard deviation since it divides by n - 1. For the normal standard deviation you need to divide by n instead.

This uses Welford's method which has higher numerical accuracy compared to the Average(x^2)-Average(x)^2 method.

Answer 3

This converts David Clarke's answer into an extension that follows the same form as the other aggregate LINQ functions like Average.

Usage would be: 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));