C# 如何确定一组值的标准偏差(STDEV)?
我需要知道与一组数字相比,一个数字是否超出平均值的1 stddev,等等。通过累积平均值和均方值,可以避免对数据进行两次传递C# 如何确定一组值的标准偏差(STDEV)?,c#,math,statistics,numerical,C#,Math,Statistics,Numerical,我需要知道与一组数字相比,一个数字是否超出平均值的1 stddev,等等。通过累积平均值和均方值,可以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 形成 sigma = sqrt(meansqr - mean^2) cnt/(cnt-1)的系数通常也是合适的 顺便
cnt = 0
mean = 0
meansqr = 0
loop over array
cnt++
mean += value
meansqr += value*value
mean /= cnt
meansqr /= cnt
形成
sigma = sqrt(meansqr - mean^2)
cnt/(cnt-1)
的系数通常也是合适的
顺便说一句——在调用Average
时,第一次传递中的数据和答案是隐藏的。在一个小的列表中,这种事情当然是微不足道的,但是如果列表超过了缓存的大小,甚至超过了工作集的大小,这就变成了一个出价交易。代码片段:
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));
}
公共静态双标准偏差(列表值列表)
{
如果(valueList.Count<2)返回0.0;
双平方和=0.0;
双平均值=valueList.average();/.NET 3.0
foreach(值列表中的双值)
{
sumOfSquares+=数学功率((值-平均值),2);
}
返回Math.Sqrt(sumOfSquares/(valueList.Count-1));
}
虽然平方和算法在大多数情况下运行良好,但如果处理的是非常大的数字,它可能会带来很大的麻烦。基本上你可能会得到一个负的方差
另外,不要,永远,永远,永远,计算a^2为pow(a,2),a*a几乎肯定更快
到目前为止,计算标准偏差的最佳方法是。我的C非常生锈,但它可能看起来像:
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));
}
公共静态双标准偏差(列表值列表)
{
双M=0.0;
双S=0.0;
int k=1;
foreach(值列表中的双值)
{
双tmpM=M;
M+=(值-tmpM)/k;
S+=(值-tmpM)*(值-M);
k++;
}
返回数学Sqrt(S/(k-2));
}
如果您拥有整个总体(与样本总体相反),那么使用return Math.Sqrt(S/(k-1))代码>
编辑:我已经根据Jason的评论更新了代码
/// <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));
}
编辑:我还根据Alex的评论更新了代码…//
/// <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));
}
///计算标准偏差,与MATLAB std(X,0)函数相同
///
///
///可枚举数据
///标准差
公共静态双GetStandardDeviation(此IEnumerable值)
{
//验证
如果(值==null)
抛出新ArgumentNullException();
int lenght=values.Count();
//从除法中节省0
如果(长度==0 | |长度==1)
返回0;
双和=0.0,sum2=0.0;
对于(int i=0;i
我发现Rob有用的答案与我在excel中看到的不太相符。为了匹配excel,我将valueList的平均值传递到标准偏差计算中
这是我的两分钱。。。很明显,你可以从函数中的valueList中计算移动平均值(ma)——但在需要标准偏差之前,我碰巧已经计算过了
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));
}
公共双标准偏差(列表值列表,双ma)
{
双xMinusMovAvg=0.0;
双西格玛=0.0;
int k=valueList.Count;
foreach(值列表中的双值){
xMinusMovAvg=值-ma;
西格玛=西格玛+(xMinusMovAvg*xMinusMovAvg);
}
返回数学Sqrt(Sigma/(k-1));
}
Jaime接受的答案很好,除了最后一行需要除以k-2(需要除以“元素的数量-1”)。
更好的是,从0开始计算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));
}
公共静态双标准偏差(列表值列表)
{
双M=0.0;
双S=0.0;
int k=0;
foreach(值列表中的双值)
{
k++;
双tmpM=M;
M+=(值-tmpM)/k;
S+=(值-tmpM)*(值-M);
}
返回数学Sqrt(S/(k-1));
}
带有扩展方法的
使用系统;
使用System.Collections.Generic;
名称空间SampleApp
{
内部课程计划
{
私有静态void Main()
{
列表数据=新列表{1,2,3,4,5,6};
双平均值=数据。平均值();
双方差=data.variance();
双sd=数据。标准偏差();
WriteLine(“均值:{0},方差:{1},标准差:{2}”,均值,方差,标准差);
Console.WriteLine(“按任意键继续…”);
Console.ReadKey();
}
}
公共静态类MyListExtensions
{
公共静态双平均值(此列表值)
{
返回值.Count==0?0:values.Mean(0,values.Count);
}
公共静态双平均值(此列表值,int开始,int结束)
{
双s=0;
for(int i=start;i0)n-=1;
收益方差/(n);
}
公共静态双标准偏差(此列表值)
{
返回值.Count==0?0:values.StandardDeviation(0,values.Count);
}
公共静态双标准偏差(此列表值,int开始,int结束)
{
双平均值=值。平均值(开始、结束);
双方差=值。方差(平均值、开始值、结束值);
var populationStdDev = new List<double>(1d, 2d, 3d, 4d, 5d).PopulationStandardDeviation();
var sampleStdDev = new List<double>(2d, 3d, 4d).StandardDeviation();
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(); }
}
private double calculateStdDev(List<double> values)
{
double average = values.Average();
return Math.Sqrt((values.Select(val => (val - average) * (val - average)).Sum()) / values.Count);
}