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Can't you just use Stdev(Odn,20)
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Cheers
Graham
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On 2/24/06, rhoemke <robert@xxxxxxxxx> wrote:
> Hello,
> i am working on this problem since quite a while.
>
> I got a special array that finds the last 30 figures of daily O-L when
> a certain condition (Open<= Pivot) was given.
> I calculate a sort of median (ODn) from these 30 figures for every day.
>
> Now i am trying to get the Standard Deviation of the last 20 ODn's,
> but i can't get this done.
> Anybody who can help?
>
> Here the code until now.
>
>
> function sort(inlist, period)
> {
> //sort inlist
> temp=0;
> for(i = period; i>0; i--)
> {
> for (j = 1; j <= i; j++)
> {
> if (inlist[j-1] > inlist[j])
> {
> temp = inlist[j-1];
> inlist[j-1] = inlist[j];
> inlist[j] = temp;
> }
> }
> }
> //inlist now sorted
> return inlist;
> }
>
>
>
>
> // MPP is a special Pivotpoint i calculate
>
> for(i=1; i<=30;i++){
> PreODn[i]= LastValue(IIf(dailyOpen <= MPP, ValueWhen(dailyopen <=
> MPP, dailyOpen - dailyLow, i) , ValueWhen(dailyopen > MPP,
> dailyOpen-dailyLow, i)));
> }
> ArrayODn = sort(PreODn, 30);
>
>
> // ODn = MA of the 30 ArrayODn but without 5 biggest and 5 smallest
> ArrayODn
> ODn = 0;
> for(i = 6; i <= 25; i++){
> ODn = ODn + ArrayODn[i];
> }
> ODn = ODn/20;
>
> //How do i get the StDev of ODn?
> // This one doesn't work :(
>
> /*for (y=BarCount-100;y<BarCount;y++){
> for (i=6; i<=25;i++){
> Mean = Sum(ODn,20)/20;
> Dev = Mean - ArrayODn[25];
> DevSq = Dev * Dev;
> MeanDevSq = Sum(DevSq, 20)/20;
> StODn = sqrt(MeanDevSq);
> }
> //}
> */
>
>
>
> Regards
> Robert
>
>
>
>
>
>
>
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