You could slow down the faster of the two.
On 4/3/06, Sashikanth Chandrasekaran <[EMAIL PROTECTED]> wrote:
...
> Assume that I do not know the distributions of p1, p2
> in advance. Is there a way I can get consistent avg.
> performance (i.e. 6!:2 'p1 V p2' = approx. 6 seconds
> and 6!:2 'p2 V p1' = approx. 6 seconds).
>
Seriously, this seems like a case where knowing more context would help.
For instance, to take a simplified version of your problem, if we have
p1=. 1 2 3 4
p2=. 2 3 4 5 6
and we want to find
(p1 e. p2);p2 e. p1
+-------+---------+
|0 1 1 1|1 1 1 0 0|
+-------+---------+
something like
r3=. 0,2=/\(ord=. \:p1,p2){p1,p2
r3,ord,(ord>#p1),:ord<:#p1
0 0 0 1 0 1 0 1 0
8 7 3 6 2 5 1 4 0
1 1 0 1 0 1 0 0 0
0 0 1 0 1 0 1 1 1
has all the data to construct an answer but may run much more quickly.
E.G. construct large versions of p1 and p2 with distributions similar to
Sashi's example:
nn=. 999
#~.p2=. 48730$~.(<?nn$0){~&.>?&.>(([: >: [: ? ])&.>50000$100)$&.><nn
48730
#~.p1=. 13930$(14000?48980){p2,(<?nn$0){~&.>?&.>(([: >: [: ?
])&.>250$100)$&.><nn
13930
NB. Arbitrary numbers above are: 48980 = 48730+250 and 0.99489588 =
48730%48980
(%/,])(6!:2 'r1 =. p1 e. p2'),6!:2 'r2 =. p2 e. p1'
2.2350347 2.7715173 1.2400332 NB. Ratio of timings, timings for r1 and
r2
6!:2 'r3=. 0,2=/\(ord=. \:p1,p2){p1,p2' NB. Faster way to get
equivalent info?
0.27696538
((+/r1) % #p1),(+/r2) % #p2
0.99475951 0.28436282 NB. p1 is 99.47% the same as p2 which is 28.4% the
same as p1
Good luck,
Devon
--
Devon McCormick
^me^ at acm.
org is my
preferred e-mail
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