These seem simpler and are possibly quicker, at least in J701 on this oldish
iPad:
100 (1{I.@E.) A NB. Fails if there’s no (second) 100 in A
719
100 ({.@}.@(I.@E.)) A NB. Returns 0 in that case
719
Easy to correct for such errors, of course.
I tried
A =. ?1000000#1000
find2 =: 13 : 'F + ((F=.>:y i. x) }. y) i. x' NB. No direct defs in J701
ts'100 find2 A'
0.020555 8.39091e6
ts'100 (1{I.@E.) A'
0.011503 76160
ts'100 ({.@}.@(I.@E.)) A'
0.007626 84864
Speed a bit better, space quite a lot, if happy with time & space tests.
Only you know if it’s wise to overwrite the first occurrence of V in A!
Cheers,
Mike
Sent from my iPad
> On 26 Aug 2022, at 19:36, Raul Miller <[email protected]> wrote:
>
> i. returns the index of the first occurrence of a value within an array.
>
> So, when implementing an algorithm which needs the index of the second
> occurrence of the value within a (large) array, we need to do some
> additional work.
>
> let's say that our array is A, the value is V
>
> F=: A i. V NB. the index of the first occurrence of V
>
> What's the most efficient way of finding the second occurrence?
>
> One possibility is
> S=: (1+F) + ((1+F) }. A) i. V
>
> Another possibility, assuming that V is numeric and not zero, would be
> S=: (0 F} A) i. V
>
> But A is large, so perhaps a faster approach would be:
> S=: {{ while. V~:y{A do. y=. y+1 end. y }} F
>
> (Which has me wishing that S=: A i.!.F V would do the job, though I'm
> not sure that that's completely appropriate...)
>
> But, we can probably eliminate the need to generate a copy of A with a
> little extra work:
>
> A=: 0 F} A
> S=: A i. V
> A=: V F} A
>
> It seems to me that this is probably going to be the fastest approach.
>
> Can anyone think of a faster approach (or something with comparable
> speed which isn't quite so unwieldy?)
>
> Thanks,
>
> --
> Raul
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