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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@xxx> 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
> > >> > >>
> > >> >
> > >>
> > >
> > >
> > >
> > > ------------------------------------
> > >
> > > Please note that this group is for discussion between users only.
> > >
> > > To get support from AmiBroker please send an e-mail directly to
> > > SUPPORT {at} amibroker.com
> > >
> > > For NEW RELEASE ANNOUNCEMENTS and other news always check DEVLOG:
> > > http://www.amibroker.com/devlog/
> > >
> > > For other support material please check also:
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> > > Yahoo! Groups Links
> > >
> > >
> > >
> >
>
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