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In answer to your question Brian, I ended up using a loop.
Seeing the power of AFL, I often find myself resisting the use of
loops, but then wrestling with the neccessary code to achieve my goals
without using a loop.
This seems to be a problem I hit quite consistently... which is a
general uncertainty about whether I can achieve what I want without a
loop or not. As it goes I usually waste time in that minor feedback
loop for a while!!!
--- In amibroker@xxxxxxxxxxxxxxx, "brianw468" <wild21@xxx> wrote:
>
> Thanks "GP", but I think your latest contribution muddies the waters
> a bit. My understanding of the original question is that varx would
> always be boolean - ie 1 or 0 (True or False)whereas your example
> treats it as an integer variable.
> The issue revolves around the real question - ie is the questioner
> asking if a single line statement can contain a recursive element
> (answer is probably NO) - or is the aim to solve a particular coding
> problem without using a loop, where the answer could well be that
> there are work-arounds. The guy who started this thread should
> clarify what he is trying to do. Otherwise, further discussion is a
> bit pointless.
>
> Brian
>
> --- In amibroker@xxxxxxxxxxxxxxx, "gp_sydney" <gp.investment@>
> wrote:
> >
> > Brian,
> >
> > As you say, it depends on what the original intention was. There are
> > two ways to interpret a statement like:
> >
> > varx = Ref(varx,-1) + 1;
> >
> > The first is the way it actually works now, where each element in
> the
> > new varx is the previous element in the old varx incremented by one.
> > So if the original varx array has:
> >
> > varx[10] = 1
> > varx[11] = 27
> > varx[12] = 39
> > varx[13] = 102
> >
> > The new varx array will have:
> >
> > varx[11] = 2
> > varx[12] = 28
> > varx[13] = 40
> > varx[14] = 103
> >
> > The other way, which is what I was talking about (as I thought it
> was
> > what was being asked about), is where the value at each bar is
> updated
> > iteratively bar by bar as it would be in a loop. So if we assume
> that
> > varx[10] is still one (but it would depend on what came before),
> then
> > we would end up with:
> >
> > varx[11] = 2 (1+1)
> > varx[12] = 3 (2+1)
> > varx[13] = 4 (3+1)
> > varx[14] = 5 (4+1)
> >
> > This is equivalent to the loop code:
> >
> > varx[i] = varx[i-1] + 1;
> >
> > Regards,
> > GP
> >
> >
> > --- In amibroker@xxxxxxxxxxxxxxx, "brianw468" <wild21@> wrote:
> > >
> > > Hi,
> > > Can you not solve the problem by (effectively) re-defining varx
> > > within the expression i.e.
> > >
> > > varx = C<Ref(L,-6) AND vary <6 AND NOT(C<Ref(L,-12) AND vary <6);
> > >
> > > Haven't tried this and the presentation could possibly be tidied
> a
> > > bit. (Depending on what, exactly, you are trying to achieve, the
> very
> > > last term might need to be "Ref(vary,-6)<6" or some such.
> > >
> > > Brian
> > >
> > >
> > > --- In amibroker@xxxxxxxxxxxxxxx, "Tomasz Janeczko" <groups@>
> > > wrote:
> > > >
> > > > It depends. The loop is general-purpose solution and works
> always.
> > > >
> > > > In some cases loops can be eliminated using Cum(), ValueWhen(),
> > > > AMA, AMA2.
> > > >
> > > > Best regards,
> > > > Tomasz Janeczko
> > > > amibroker.com
> > > > ----- Original Message -----
> > > > From: "sidhartha70" <sidhartha70@>
> > > > To: <amibroker@xxxxxxxxxxxxxxx>
> > > > Sent: Thursday, September 18, 2008 9:56 AM
> > > > Subject: [amibroker] Re: Recursive Boolean Expressions...
> Possible?
> > > >
> > > >
> > > > > Can I ask the master...?? TJ... Does this kind of expression
> > > > > absolutely require a loop structure?
> > > > >
> > > > > TIA
> > > > >
> > > > > --- In amibroker@xxxxxxxxxxxxxxx, "gp_sydney"
> <gp.investment@>
> > > wrote:
> > > > >>
> > > > >> Graham,
> > > > >>
> > > > >> That doesn't work either, in the general case, as varx is
> still
> > > not
> > > > >> dependent on previous values of varx, only on previous
> values of
> > > your
> > > > >> first "temp" statement.
> > > > >>
> > > > >> Consider the simpler case:
> > > > >>
> > > > >> temp = BarIndex() < 10;
> > > > >> varx = temp AND NOT Ref(temp,-1);
> > > > >>
> > > > >> temp now has the first 10 bars set to one and all other bars
> set
> > > to
> > > > >> zero. varx will have the first 11 bars set to zero, since Ref
> > > (temp,-1)
> > > > >> is one (actually the first bar will probably be null) and
> then
> > > all
> > > > >> subsequent bars will also be zero since temp is then zero.
> > > > >> Consequently, varx would be completely zero, except perhaps
> for
> > > the
> > > > >> first null.
> > > > >>
> > > > >> Assuming this did work as suggested, compare to:
> > > > >>
> > > > >> varx = BarIndex() < 10 AND NOT Ref(varx,-1);
> > > > >>
> > > > >> Actually if the first bar was null due to Ref(varx,-1) being
> > > null,
> > > > >> then varx would end up completely full of nulls (a problem
> to be
> > > wary
> > > > >> of with nulls in loops). But say the first bar ended up
> being
> > > zero
> > > > >> (perhaps the nz function was used), then the second bar
> would be
> > > one,
> > > > >> since BarIndex is less than 10 and Ref(varx,-1) refers to
> the
> > > first
> > > > >> bar which we just said was zero. The third bar would be
> zero,
> > > since
> > > > >> Ref(varx,-1) now refers to the second bar which we just set
> to
> > > one,
> > > > >> and the fourth bar would be one again. This would continue
> up to
> > > the
> > > > >> 10th bar, after which all bars would be zero due to the
> BarIndex
> > > term.
> > > > >> The first 10 bars of varx alternating between one and zero
> make
> > > the
> > > > >> result different to the first version.
> > > > >>
> > > > >> Regards,
> > > > >> GP
> > > > >>
> > > > >>
> > > > >> --- In amibroker@xxxxxxxxxxxxxxx, Graham <kavemanperth@>
> wrote:
> > > > >> >
> > > > >> > try this
> > > > >> > temp = C<Ref(L,-6) AND vary<6;
> > > > >> > varx = temp AND NOT Ref(temp ,-6);
> > > > >> >
> > > > >> > --
> > > > >> > Cheers
> > > > >> > Graham Kav
> > > > >> > AFL Writing Service
> > > > >> > http://www.aflwriting.com
> > > > >> >
> > > > >> >
> > > > >> >
> > > > >> >
> > > > >> > 2008/9/18 gp_sydney <gp.investment@>:
> > > > >> > > No, you can't do that as the right-hand expression is
> > > evaluated
> > > > > on the
> > > > >> > > whole array before anything is assigned to the left-hand
> > > variable.
> > > > >> > > That means that "varx" is effectively constant during
> the
> > > expression
> > > > >> > > evaluation for the whole array. It's essentially the
> same as:
> > > > >> > >
> > > > >> > > temp = IIf(C<Ref(L,-6) AND vary<6 AND NOT Ref(varx,-
> > > 6),True,False);
> > > > >> > > varx = temp;
> > > > >> > >
> > > > >> > > To do what you are suggesting would require a loop.
> > > > >> > >
> > > > >> > > Regards,
> > > > >> > > GP
> > > > >> > >
> > > > >> > >
> > > > >> > > --- In amibroker@xxxxxxxxxxxxxxx, "sidhartha70"
> > > <sidhartha70@>
> > > > > wrote:
> > > > >> > >>
> > > > >> > >> Hi All,
> > > > >> > >>
> > > > >> > >> Is it possible to have recursive boolean
> expressions...?
> > > i.e. the
> > > > >> true
> > > > >> > >> or false of the current value of the array depends on
> > > whether a
> > > > >> > >> previous value of the array is true or false.
> > > > >> > >>
> > > > >> > >> So for example,
> > > > >> > >>
> > > > >> > >> varx = IIf(C<Ref(L,-6) AND vary<6 AND NOT Ref(varx,-
> > > 6),True,False);
> > > > >> > >>
> > > > >> > >> Would that work... or are recursive booleans like this
> not
> > > > > allowed??
> > > > >> > >>
> > > > >> > >> TIA
> > > > >> > >>
> > > > >> >
> > > > >>
> > > > >
> > > > >
> > > > >
> > > > > ------------------------------------
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> only.
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> > > > >
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> > >
> >
>
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