> From: "Sherlock, Ric" <[EMAIL PROTECTED]>
>
> ---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
For completeness,
f3=: (,:"0~ 0,+/[EMAIL PROTECTED]:)@[ <;.0 ]
x (f0 -: f2) y
1
3 ts 'x f1 y'
0.00764483 4.21952e6
3 ts 'x f3 y'
0.0122254 5.05888e6
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