Should be shorter
> movingAvg =: dyad define
> (+/%#) (x*y) , }:L
> )
I don't think it's possible to get rid of L.
How would you know what to }: each time?
--- Dan Bron <[EMAIL PROTECTED]> wrote:
> I need to "undo" an average. The problem is I don't have the whole series of
> data; I have only
> the initial average and, periodically, a new value to be averaged in.
>
> So, let's say a user hands me a pre-computed average. I want to take that
> average, and whenever
> I get a new value of the series, I want to "expand" the precomputed average,
> drop off its
> leading (oldest) value, append the new value, and redo the average. Then I
> would use this new
> average as my "precomputed" average.
>
> One wrinkle is that the average is weighted. In particular, the newest value
> is always weighted
> by some user-specified factor (all other weights are 1). The initial
> user-provided precomputed
> average has already taken this into account for the first iteration, and my
> code is responsible
> for taking it into account for any new values.
>
>
> Concretely, I would like to emulate this J verb:
>
> L =: ?.~10 NB. Original list -- I won't have access to
> this.
> wmr =: 3 NB. Weight Most Recent value by a factor of 3
>
> movingAvg =: dyad define
> (+/%#) (x*y) , }:L
> )
>
> wmr movingAvg 10
> 7
> wmr movingAvg 5
> 5.8
> wmr movingAvg 8
> 6.4
> wmr movingAvg 9
> 7.2
>
>
> except that I do not have access to the original series of data L . In my
> scenario, my
> initial inputs would be wmr and (+/%#) (wmr*{.L) , }.L and, later, the
> values 10 5 8 9
> (spread out over time). Ideally, I would like to avoid keeping any kind of
> list (e.g., of the
> new values as they come in).
>
> Is this possible?
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