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Thanks for you help
I do not have much of a code....at least not for this segment
Median is found using the standard AFL function
I can now envisage the following pseudocode
medd = median(c,260)
for k = 1 to 260
medo [k] = medd
next
freq = 0
for j = 1 to 260
if c[i] = medo[i] then feq = freq +1
next
addcolumn frequency
Thanks for pointing the = and ==
--- In amibroker@xxxxxxxxxxxxxxx, "gp_sydney" <gp.investment@xxx>
wrote:
>
> > if c [i] = median (close,60)
>
> That won't work because you're trying to compare a number with an
> array of numbers, plus you need the relational equals operator (==)
> rather than the assignment one (=). Read the median array into a
> variable first and then use indexing on that:
>
> medc = median(close,60);
> if (c[i] == medc[i])
>
> However, without seeing the rest of your code, I have a feeling you
> still won't end up with what you're expecting. This is comparing
the
> current bar's close with the median of the previous 60 bars
(counting
> the current one), so if you only count those, it will give you a
count
> of how many bars had a close equal to the median of the last 60
bars.
> If at each bar you want a count of how many bars in the previous 60
> were the same as the median over that range, then the calculation
will
> be different, probably involving nested loops.
>
> GP
>
>
> --- In amibroker@xxxxxxxxxxxxxxx, "meenhal" <meenhal@> wrote:
> >
> > To set an entry point, I am trying to find price deviations from
mean
> > and median for different intervals.
> >
> > All is easy except: I can not determine the frequecy of median
i.e.
> > the number of occurences when the price was median during the
inteval.
> >
> > median (close,260) will give me the median closing price for 260
bars
> > but it does not tell whether this prices occured 10 tims or 88
times.
> >
> > I tried to use the for loop with if c [i] = median (close,60)
but
> > AFL's handling of arrays does not permit such an attepmt.
> >
> > Can anyone please help. Thanks in advance
> >
>
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