Question

我需要知道一组数字与一组数字相比是否超出平均值的1 stddev等。

Answer 1

虽然平方和算法在大多数情况下都能正常工作，但是如果处理非常大的数字会导致大麻烦。你基本上可能会出现负差异......

另外，永远不要永远计算^ 2作为pow（a，2），a * a几乎肯定会更快。

到目前为止，计算标准差的最佳方法是Welford's method。我的C非常生疏，但看起来像是：

public static double StandardDeviation(List<double> valueList)
{
    double M = 0.0;
    double S = 0.0;
    int k = 1;
    foreach (double value in valueList) 
    {
        double tmpM = M;
        M += (value - tmpM) / k;
        S += (value - tmpM) * (value - M);
        k++;
    }
    return Math.Sqrt(S / (k-2));
}

如果您拥有整个人口（而不是示例人口），请使用return Math.Sqrt(S / (k-1));。

编辑：我根据Jason的评论更新了代码......

编辑：我还根据Alex的评论更新了代码......

Answer 2

比Jaime的

快10倍解决方案，但要注意，正如杰米所指出的那样：

＆＃34;虽然平方和算法在大多数情况下工作正常，但它如果您正在处理非常大的数字，可能会造成大麻烦。您基本上可能会出现负差异＆＃34;

如果你认为你正在处理非常大的数字或非常大量的数字，你应该使用这两种方法计算，如果结果相同，你肯定知道你可以使用＆＃34;我的＆＃34;你的方法。

    public static double StandardDeviation(double[] data)
    {
        double stdDev = 0;
        double sumAll = 0;
        double sumAllQ = 0;

        //Sum of x and sum of x²
        for (int i = 0; i < data.Length; i++)
        {
            double x = data[i];
            sumAll += x;
            sumAllQ += x * x;
        }

        //Mean (not used here)
        //double mean = 0;
        //mean = sumAll / (double)data.Length;

        //Standard deviation
        stdDev = System.Math.Sqrt(
            (sumAllQ -
            (sumAll * sumAll) / data.Length) *
            (1.0d / (data.Length - 1))
            );

        return stdDev;
    }

Answer 3

Math.NET库为您提供了这个功能。

PM＆GT;安装包MathNet.Numerics

var populationStdDev = new List<double>(1d, 2d, 3d, 4d, 5d).PopulationStandardDeviation();

var sampleStdDev = new List<double>(2d, 3d, 4d).StandardDeviation();

有关详细信息，请参阅http://numerics.mathdotnet.com/docs/DescriptiveStatistics.html。

Answer 4

Jaime接受的答案很棒，除非您需要在最后一行中除以k-2（您需要除以“number_of_elements-1”）。更好的是，从0开始k：

public static double StandardDeviation(List<double> valueList)
{
    double M = 0.0;
    double S = 0.0;
    int k = 0;
    foreach (double value in valueList) 
    {
        k++;
        double tmpM = M;
        M += (value - tmpM) / k;
        S += (value - tmpM) * (value - M);
    }
    return Math.Sqrt(S / (k-1));
}

Answer 5

您可以通过累积均值和均方

来避免对数据进行两次传递

cnt = 0
mean = 0
meansqr = 0
loop over array
    cnt++
    mean += value
    meansqr += value*value
mean /= cnt
meansqr /= cnt

并形成

sigma = sqrt(meansqr - mean^2)

因素cnt/(cnt-1)通常也是合适的。

BTW--对Average和Demi答案中数据的第一次传递隐藏在对{{1}}的调用中。这样的事情在一个小清单上肯定是微不足道的，但如果列表超过缓存的大小，甚至是工作集，那么这就是买卖协议。

Answer 6

代码段：

public static double StandardDeviation(List<double> valueList)
{
    if (valueList.Count < 2) return 0.0;
    double sumOfSquares = 0.0;
    double average = valueList.Average(); //.NET 3.0
    foreach (double value in valueList) 
    {
        sumOfSquares += Math.Pow((value - average), 2);
    }
    return Math.Sqrt(sumOfSquares / (valueList.Count - 1));
}

Answer 7

我发现Rob的有用答案与我使用excel看到的完全不一致。为了匹配excel，我将AverageList中的Average传递给StandardDeviation计算。

这是我的两分钱......显然你可以从函数里面的valueList计算移动平均线（ma） - 但是在我需要standardDeviation之前我就已经知道了。

public double StandardDeviation(List<double> valueList, double ma)
{
   double xMinusMovAvg = 0.0;
   double Sigma = 0.0;
   int k = valueList.Count;


  foreach (double value in valueList){
     xMinusMovAvg = value - ma;
     Sigma = Sigma + (xMinusMovAvg * xMinusMovAvg);
  }
  return Math.Sqrt(Sigma / (k - 1));
}

Answer 8

使用扩展方法。

using System;
using System.Collections.Generic;

namespace SampleApp
{
    internal class Program
    {
        private static void Main()
        {
            List<double> data = new List<double> {1, 2, 3, 4, 5, 6};

            double mean = data.Mean();
            double variance = data.Variance();
            double sd = data.StandardDeviation();

            Console.WriteLine("Mean: {0}, Variance: {1}, SD: {2}", mean, variance, sd);
            Console.WriteLine("Press any key to continue...");
            Console.ReadKey();
        }
    }

    public static class MyListExtensions
    {
        public static double Mean(this List<double> values)
        {
            return values.Count == 0 ? 0 : values.Mean(0, values.Count);
        }

        public static double Mean(this List<double> values, int start, int end)
        {
            double s = 0;

            for (int i = start; i < end; i++)
            {
                s += values[i];
            }

            return s / (end - start);
        }

        public static double Variance(this List<double> values)
        {
            return values.Variance(values.Mean(), 0, values.Count);
        }

        public static double Variance(this List<double> values, double mean)
        {
            return values.Variance(mean, 0, values.Count);
        }

        public static double Variance(this List<double> values, double mean, int start, int end)
        {
            double variance = 0;

            for (int i = start; i < end; i++)
            {
                variance += Math.Pow((values[i] - mean), 2);
            }

            int n = end - start;
            if (start > 0) n -= 1;

            return variance / (n);
        }

        public static double StandardDeviation(this List<double> values)
        {
            return values.Count == 0 ? 0 : values.StandardDeviation(0, values.Count);
        }

        public static double StandardDeviation(this List<double> values, int start, int end)
        {
            double mean = values.Mean(start, end);
            double variance = values.Variance(mean, start, end);

            return Math.Sqrt(variance);
        }
    }
}

Answer 9

/// <summary>
/// Calculates standard deviation, same as MATLAB std(X,0) function
/// <seealso cref="http://www.mathworks.co.uk/help/techdoc/ref/std.html"/>
/// </summary>
/// <param name="values">enumumerable data</param>
/// <returns>Standard deviation</returns>
public static double GetStandardDeviation(this IEnumerable<double> values)
{
    //validation
    if (values == null)
        throw new ArgumentNullException();

    int lenght = values.Count();

    //saves from devision by 0
    if (lenght == 0 || lenght == 1)
        return 0;

    double sum = 0.0, sum2 = 0.0;

    for (int i = 0; i < lenght; i++)
    {
        double item = values.ElementAt(i);
        sum += item;
        sum2 += item * item;
    }

    return Math.Sqrt((sum2 - sum * sum / lenght) / (lenght - 1));
}

Answer 10

所有其他答案的问题在于他们认为你拥有自己的答案大数据中的数据。如果你的数据是即时进入的，那就是更好的方法。无论您如何存储数据，此类都有效。它还为您提供Waldorf方法或平方和方法的选择。两种方法都使用单次传递。

public final class StatMeasure {
  private StatMeasure() {}

  public interface Stats1D {

    /** Add a value to the population */
    void addValue(double value);

    /** Get the mean of all the added values */
    double getMean();

    /** Get the standard deviation from a sample of the population. */
    double getStDevSample();

    /** Gets the standard deviation for the entire population. */
    double getStDevPopulation();
  }

  private static class WaldorfPopulation implements Stats1D {
    private double mean = 0.0;
    private double sSum = 0.0;
    private int count = 0;

    @Override
    public void addValue(double value) {
      double tmpMean = mean;
      double delta = value - tmpMean;
      mean += delta / ++count;
      sSum += delta * (value - mean);
    }

    @Override
    public double getMean() { return mean; }

    @Override
    public double getStDevSample() { return Math.sqrt(sSum / (count - 1)); }

    @Override
    public double getStDevPopulation() { return Math.sqrt(sSum / (count)); }
  }

  private static class StandardPopulation implements Stats1D {
    private double sum = 0.0;
    private double sumOfSquares = 0.0;
    private int count = 0;

    @Override
    public void addValue(double value) {
      sum += value;
      sumOfSquares += value * value;
      count++;
    }

    @Override
    public double getMean() { return sum / count; }

    @Override
    public double getStDevSample() {
      return (float) Math.sqrt((sumOfSquares - ((sum * sum) / count)) / (count - 1));
    }

    @Override
    public double getStDevPopulation() {
      return (float) Math.sqrt((sumOfSquares - ((sum * sum) / count)) / count);
    }
  }

  /**
   * Returns a way to measure a population of data using Waldorf's method.
   * This method is better if your population or values are so large that
   * the sum of x-squared may overflow. It's also probably faster if you
   * need to recalculate the mean and standard deviation continuously,
   * for example, if you are continually updating a graphic of the data as
   * it flows in.
   *
   * @return A Stats1D object that uses Waldorf's method.
   */
  public static Stats1D getWaldorfStats() { return new WaldorfPopulation(); }

  /**
   * Return a way to measure the population of data using the sum-of-squares
   * method. This is probably faster than Waldorf's method, but runs the
   * risk of data overflow.
   *
   * @return A Stats1D object that uses the sum-of-squares method
   */
  public static Stats1D getSumOfSquaresStats() { return new StandardPopulation(); }
}

Answer 11

我们也许可以在Python中使用统计信息模块。它具有stedev（）和pstdev（）命令，分别计算样本和总体的标准偏差。

详细信息：https://www.geeksforgeeks.org/python-statistics-stdev/

将统计信息导入为st print（st.ptdev（dataframe ['column name']））

Answer 12

这是人口标准差

private double calculateStdDev(List<double> values)
{
    double average = values.Average();
    return Math.Sqrt((values.Select(val => (val - average) * (val - average)).Sum()) / values.Count);
}

对于样本标准偏差，只需在上述代码中将[values.Count]更改为[values.Count -1]。

确保您的集合中没有只有1个数据点。

如何确定一组值的标准偏差（stddev）？

12 个答案: