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I thought he was after an xaverage of the 20 elements in his array, not a 20 bar xaverage of the [0] element of the array... Your code gives him the xaverage of the [0] element over 20 bars and ignores the other 19 elements of the array on each bar...
---- you wrote:
> So all you really need is an xaverage of myarray[0].
>
> Try something like...
>
> fact = 2/(len+1);
> av = (1-fact)*av[1] + fact*myarray[0];
>
> > No ! My array indexes in the new value at 0 and kicks out the last at 20 and
> > moves everything down at the close of each bar. I am sure it can be done, it
> > is just a little over my head. Thanks for your input.
>
> --
> Dennis
>
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