At the same time it was puzzling how the most
straightforward method--passing concrete indices
and lengths--did not produce a good speed.
f3=: (,:"0~ 0,+/[EMAIL PROTECTED]:)@[ <;.0 ]
f3a=: <;.0
z=. (,:"0~ 0,+/[EMAIL PROTECTED]:) x
x (f0 -: f3) y
1
z ((x f0 ]) -: f3a) y
1
3 ts 'x f0 y'
0.0341443 1.09757e7
3 ts 'x f1 y'
0.00830281 4.22656e6
3 ts 'x f3 y'
0.0155314 5.07117e6
3 ts 'z f3a y'
0.0116746 4.94016e6
> From: Roger Hui <[EMAIL PROTECTED]>
>
> It has been a long quest in APL/J to make
> x e. y as fast as 1 x1}y1 . The quest continues.
>
> I must say that I like Brian Schott's solution (I.x)> the best. The
> implementation should reward concise
> solutions and I'll see if I can make that solution faster.
>
>
> ----- Original Message -----
> From: "Sherlock, Ric"
> Date: Tuesday, November 4, 2008 22:57
> Subject: RE: [Jprogramming] cut on length
> To: Programming forum
>
> > ---Roger Hui wrote:
> > > f0=: 4 : '(I.x)> > > f1=: 4 : '((i.#y) e. +/\0,x) <;.1 y'
> > >
> > > x=: 1+1e4 [EMAIL PROTECTED] 100
> > > y=: t {~ (+/x) [EMAIL PROTECTED] #t=. 'barack obama'
> > >
> > > x (f0 -: f1) y
> > > 1
> > > ts=: 6!:2 , 7!:[EMAIL PROTECTED]
> > >
> > > ts 'x f0 y'
> > > 0.0700156 1.09591e7
> > > ts 'x f1 y'
> > > 0.00778395 4.20992e6
> >
> > When I saw Brian's solution I wasn't going to post mine - his
> > was so much simpler (BTW I have only used monadic I. to turn a
> > boolean into an index of 1s and hadn't realised what it did with
> > integers). Roger's post prompted me to time mine and the result
> > surprised me.
> >
> > f2=: 4 : '(1 (<:+/\x)} (#y)$0) <;.2 y'
> > x (f1 -: f2) y
> > 1
> > 50 ts 'x f0 y'
> > 0.030942459167 10976256
> > 50 ts 'x f1 y'
> > 0.0077791269561 4227072
> > 50 ts 'x f2 y'
> > 0.0060425808308 2654272
> ----------------------------------------------------------------------
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