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My dictionary defines it as the average of the two middle values when the number sequence contains an even number of values. If the sequence contains an odd number of values, the middle value would be the median.
I'm just having a heck of a time trying to figure out how to get AB to segregate the four highest and four lowest values out of the entire array in order to obtain the medians...
Any ideas?
DC
----- Original Message -----
From: kaveman perth
To: amibroker@xxxxxxxxxxxxxxx
Sent: Friday, November 05, 2004 10:40 PM
Subject: Re: [amibroker] Re: median calculation
From the way I work the median in AB, it goes like this
median(C,5) list out the last 5 bars and sorts them in numeric order.
The mid price(ie 3rd) is used as the median. In the case of say 6 bars
the 3rd is also used
for 7 & 8 bars the 4th used etc
On Fri, 5 Nov 2004 20:33:50 -0500, dennisconn@xxxxxxxxxxxxxxxx
<dennisconn@xxxxxxxxxxxxxxxx> wrote:
>
> ----- Original Message -----
> From: Bob Jagow
>
> Whoops--now you want the average instead of the median?
>
> *** I want the median of the four highest peaks, which is the same thing as the average of the second and third highest peaks, yes? ***
>
> For that, just sum peak(myOsc,chng,i) and trough(myOsc,chng, i) for the
> ith most recent peaks. [Note that the peaks and troughs are supposed to
> be > 0, so add a constant if needed.]
>
> *** I only want the median value of the four highest peaks and the median value of the four lowest troughs - how is summing all the peaks and troughs going to help find the median? I assume "chng" is short for "change"? The change from what to what? Seriously - I really am not following what you're trying to convey here... ***
>
> A bubble sort just swaps the values in 2 loops [Google bubble sort]. Do
> this with AB's for loop instead of calling VBscript.
>
> *** OK, I'll try the "for" loop solution... are you thinking nested loops? I once programmed a couple of nested for-next loops to find the median value of a fixed-length array, and even though I don't know what a Google bubble sort is, what I did sounds like a bubble sort (as I mentioned, that was 20 years ago). I guess I'll have to try to recreate it using the AB version of the next-less "for" routine (I have a hard time with the idea that you can use "for" without "next", but that's my old-dog-can't-learn-new-tricks mindset). As for VBscript, is that the same thing as Visual BASIC? If so, is VB pretty much the same as plain old generic BASIC? I might be better off just using that, since I can grasp BASIC easier than I seem to be getting AFL...***
>
> ???
>
> DC
>
> -----Original Message-----
> From: dennisconn@xxxxxxxxxxxxxxxx [mailto:dennisconn@xxxxxxxxxxxxxxxx]
> Sent: Friday, November 05, 2004 1:33 PM
> To: amibroker@xxxxxxxxxxxxxxx
> Subject: Re: [amibroker] Re: median calculation
>
> Hi Bob,
>
> Thanks for the response. The only problem is that I want to chart some
> further calculations in AB which will be dependent on the median
> calculation - so Excel won't get it.
>
> All I can think of is to run a loop which first sets the HHV of the
> array to zero, then finds the next highest value, assign it to x, set
> THAT high value in the array to zero, then look for the next high value
> and average it with x. Then do the same for the negative values in the
> array to find the median of the four lowest troughs.
>
> Even if I figure out how to do that, it seems awfully clumsy. There has
> to be a better way, but I have a feeling it's going to involve some kind
> of nested loop in VBscript or something, to sort them out. I haven't
> written anything serious in 20 years, and I've forgotten most of the
> how-to-do-it solutions I found then.
>
> While writing this, I noticed your follow-up response mentioning bubble
> sorts. I'm not familiar with the term, but I have the feeling it
> involves what I considered in the last paragraph.
>
> Still stumped, but plugging away...
>
> DC
> ----- Original Message -----
> From: Bob Jagow
> To: amibroker@xxxxxxxxxxxxxxx
> Sent: Friday, November 05, 2004 4:08 PM
> Subject: RE: [amibroker] Re: median calculation
>
> Since "The Median function - finds median (middle element) value of
> the
> array over period elements", you'd have to identify the "four highest
> peaks within a user-defined lookback range as well as the four lowest
> troughs below zero" and try to stuff them into a new array in order to
> use it.
> I'd suggest Excel in preference to AB.
>
> Regards,
> Bob
>
> -----Original Message-----
> From: conn_dennis [mailto:dennisconn@xxxxxxxxxxxxxxxx]
> Sent: Friday, November 05, 2004 12:26 PM
> To: amibroker@xxxxxxxxxxxxxxx
> Subject: [amibroker] Re: median calculation
>
> So was the question that stupid, or hasn't anyone ever attempted it?
>
> Anyone???
>
> --- In amibroker@xxxxxxxxxxxxxxx, "conn_dennis" <dennisconn@xxxx>
> wrote:
> >
> > I know that the median function will return the median value of the
> > entire array, but how would one go about finding the median of only
> > some of the values in an array?
> >
> > So far, I've managed only to identify the peaks in an oscillator
> I've
> > coded, but I'm trying to have the program find the median of ONLY
> the
> > four highest peaks within a user-defined lookback range (likewise
> > with the four lowest troughs below zero).
> >
> > I think I can stumble around and figure out how to code what I need
> > once I get past this median problem. I've spent two days on this
> > trying to figure it out (I feel like Thomas Edison - I've
> discovered
> > a thousand ways to NOT find the median). : )
> >
> > Has anyone had a similar problem and found a somewhat simple
> > solution, or perhaps be willing to point me in some direction?
> >
> > DC
>
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--
Cheers
Graham
http://e-wire.net.au/~eb_kavan/
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