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This is just to say a big thank you
To read someone else's problem and then actually write a piece of
workable code is really something extremely admirable.
I tried the code, it went on without errors................from then
on the worries are my own.
HOWEVER: for theoritical interest...........it will be nice to find
the median for the same set of data using Amibroker and the good old
Excel.
Thanks again for the help
--- In amibroker@xxxxxxxxxxxxxxx, "gp_sydney" <gp.investment@xxx>
wrote:
>
> 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@> 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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