“분산 계산하기위한 알고리즘”에 대한 위키 백과의 문서 섹션 방법 요소가 귀하의 관찰에 추가하는 경우 분산을 계산하는 방법을 보여줍니다. (표준 편차는 분산의 제곱근입니다.) x n + 1을 더 한다고 가정합니다.xn+1

을 배열에 추가 한 다음

EDIT: Above formula seems to be wrong, see comment.

Now, replacing an element means adding an observation and removing another one; both can be computed with the formula above. However, keep in mind that problems of numerical stability may ensue; the quoted article also proposes numerically stable variants.

To derive the formula by yourself, compute (n−1)(σ2new−σ2old)

using the definition of sample variance and substitute μnew

by the formula you gave when appropriate. This gives you σ2new−σ2old

in the end, and thus a formula for σnew

given σold

and μold

. In my notation, I assume you replace the element xn

by x′n

:

The xk

in the sum transform into something dependent of μold

, but you’ll have to work the equation a little bit more to derive a neat result. This should give you the general idea.

답변

real sum = 0;
int count = 0;
real S = 0;
real variance = 0;

real GetRunningStandardDeviation(ref sum, ref count, ref S, x)
{
   real oldMean;

   if (count >= 1)
   {
       real oldMean = sum / count;
       sum = sum + x;
       count = count + 1;
       real newMean = sum / count;

       S = S + (x-oldMean)*(x-newMean)
   }
   else
   {
       sum = x;
       count = 1;
       S = 0;
   }

   //estimated Variance = (S / (k-1) )
   //estimated Standard Deviation = sqrt(variance)
   if (count > 1)
      return sqrt(S / (count-1) );
   else
      return 0;
}