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Here's a function that I think will do what you want. I haven't tried
it though, so there could be mistakes.
function MedianCount(data, period)
{
medo = Median(data, period);
medCnt = 0;
for (i = 0; i < BarCount; i++)
{
cnt = period;
for (j = i; j >= 0 && cnt > 0; j--, cnt--)
{
if (data[j] == medo[i])
medCnt[i]++;
}
}
return medCnt;
}
If you only want to include periods with the full "period" bars, start
the 'i' loop at "period" rather than zero. The "j >= 0" part of the
inner loop could then be dispensed with as well.
GP
--- In amibroker@xxxxxxxxxxxxxxx, "meenhal" <meenhal@xxx> wrote:
>
>
> Your kind attention to detail must be gratefully admired
>
>
> I used the following code
>
>
> medo = Median(C,65);
>
> medoo [0] = medo;
> for( i = 1; i < 65; i++ )
> { medoo[ i ] = medo; }
>
> freq=0;
>
> for( i = 1; i < 65; i++ )
> { if ( C[i] == medoo[ i ] )
>
> freq= freq+1 ; )
>
> First loop was an attempt to populate an array with the value of
> median so that then the array [c] could be compared with it. Well it
> did not work. The only option left is to actually find the numerical
> value of the median for each ticker in the first instance and then
> mechanically plug in that number in the second loop to find the
> frequency. This seems to be the dumbest way of programming. I am
> sure there must be a better way.
>
> More help please
> Ta
>
>
>
>
>
> --- In amibroker@xxxxxxxxxxxxxxx, "gp_sydney" <gp.investment@>
> wrote:
> >
> > Firstly, loops should go from zero to last-1 rather than one to
> last.
> The first FOR loop was to populate a dummy array with > Also, if
> you're going to use an absolute value for the limit (ie. the
> > 260 in your example) then you need to add checking for that being
> out
> > of range of the current chart. Arrays are only dimensioned up to
> > BarCount, whatever that is for the current chart, so if BarCount is
> > less than 260 in this case you'll have a problem.
> >
> > Finally, I don't think that logic will give you what you want. That
> > just checks, for each bar, if that bar is the same as the median of
> > the last 260 days and then adds them up, which is what I mentioned
> > before. I gather you want a value at each bar of the number of
> bars in
> > the last interval that match the median over the same interval. I
> > think you'll need nested loops for that, one to go through the
> bars,
> > and another inside it to count back through the previous interval
> (eg.
> > 260) number of bars, with appropriate range checking.
> >
> > Regards,
> > GP
> >
> >
> > --- In amibroker@xxxxxxxxxxxxxxx, "meenhal" <meenhal@> wrote:
> > >
> > > 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@>
> > > 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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