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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...
>
> Dennis' code is probably what you really want.
I agree, Dennis has the right idea.
You can take the **average** of the elements in a 20-element array,
but OM is right that it doesn't make a lot of sense to take the
**xaverage** of the array. You could run the xaverage calculation on
just the 20 elements of the array, as JimO showed, but I doubt that's
what you want. That gives you the equivalent of an xaverage on the
first 20 bars of a chart, and discards all information about the
previous data. It certainly wouldn't give the same results as a
"real" xaverage.
average uses ONLY the data in the last N bars, which is why e.g. a
spike or gap will suddenly drop out of your average N bars later.
xaverage is an ongoing calculation that depends on the previous data
fed into it, including data more than N bars ago. That's why
xaverage doesn't have the N-bars-later-dropout problem. Instead of
dropping out abruptly like in a simple average, the spike fades out
gradually with an exponential decay. And you only need one data
point (the [0] element in your case) to compute the xaverage of the
"last N elements" -- because xaverage has already incorporated the
previous N-1 elements (and, actually, ALL previous elements) into the
previous xaverage value.
Gary
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