Edson,
I did manage to use the 'CollectSetAccumulateFunction' as a template for the
MatchAll and have also successfully 'called' the accumulate function from a
rule written in a .drl file. But I'm stuck at one point - how to send the
value of $param to the accumulate method ? From what I
Up to Drools 5, the accumulate function API supports only a single
argument to the function, but that argument can be an array or a list. So
what I do on my own implementations is to pass the $param as one element of
the array and $o as the other. It is not ideal, but it is how you can do
that
Hey again everyone,
I seem to be having another problem.
When I have a rule like that
when
A() over window:length(1)
B() over window:length(1)
then
...
I imagine now having two windows of length 1. One always keeping the latest
A event and one always keeping
Hello everyone.
Lets say we have StockTicks...
Is it possible to detect a monotonically decreasing StockTick stream over a
window?
If I knew the number of events I want to check on, I'd do it this way
$a : StockTick() over window:time(20s)
$b : StockTick(this after $a, price $a.price)
See the current discussion of accumulate functions. That should satisfy your
need.
GreG
On Jan 29, 2011, at 10:18, OlliSee o.ro...@seeburger.de wrote:
Hello everyone.
Lets say we have StockTicks...
Is it possible to detect a monotonically decreasing StockTick stream over a
window?
Is there a *defined *order in which objects resulting from an over clause
are presented to the preceding from? Otherwise, an accumulate function
requiring a specific order is based on implementation details.
-W
On 29 January 2011 18:24, Greg Barton greg_bar...@yahoo.com wrote:
See the current
Edson,
That worked beautifully !
Drools is looking more and more promising for what we need to do.
Thanks for all your help,
Gurvinder
-Original Message-
From: Edson Tirelli-4 [via Drools - Java Rules Engine]
[mailto:ml-node+2380361-572428564-9...@n3.nabble.com]
Sent: Sat
how about numbering your ticks ?
as a rough outline, this might look something like:
1st : Tick ($number : num)
2nd : Tick (num == $number+1, {this.value below 1st.value})
3rd : Tick (num == $number+2, {this.value above 2nd.value})
this should find all bottoms
--
View this message in context:
