Perhaps the left argument could better be given as #~/x as in
#~/ 1 2 3 2 3 1,:2 1 3 2 1 3
1 1 2 3 3 3 2 2 3 1 1 1
where the value gives the level depth and the value ranges
determine the ranges in y .
Then a solution is
foo1=: 4 : '<@]^:({...@[)&>/"1 |:(<1 ,~ (}: ~: }.)x) <;.2 &> x ; y'
1 1 2 3 3 3 2 2 3 1 1 1 foo1 i.12
+---+---+---------+-----+-----+-------+
|0 1|+-+|+-------+|+---+|+---+|9 10 11|
| ||2|||+-----+|||6 7|||+-+|| |
| |+-+|||3 4 5|||+---+|||8||| |
| | ||+-----+|| ||+-+|| |
| | |+-------+| |+---+| |
+---+---+---------+-----+-----+-------+
R.E. Boss
> -----Oorspronkelijk bericht-----
> Van: [email protected] [mailto:programming-
> [email protected]] Namens R.E. Boss
> Verzonden: donderdag 4 juni 2009 8:49
> Aan: 'Programming forum'
> Onderwerp: Re: [Jprogramming] leveled as indicated
>
> My solution was
>
> foo=: 4 : 0
> 'x0 x1'=. x
> z=. (<"0 x0) ,. (x1 # i.#x1)</.y
> (<@]^:[&>/)"1 z
> )
>
> It contains parts of both your and Hui's solution.
> Thanks.
>
>
> R.E. Boss
>
>
> > -----Oorspronkelijk bericht-----
> > Van: [email protected] [mailto:programming-
> > [email protected]] Namens Raul Miller
> > Verzonden: woensdag 3 juni 2009 21:42
> > Aan: Programming forum
> > Onderwerp: Re: [Jprogramming] leveled as indicated
> >
> > foo=:4 :0
> > assert. 4 = (3!:0) x NB. x is integer
> > assert. 2 = # x NB. x has 2 items
> > assert. 2 = #$ x NB. x has rank 2
> > assert. (# y) = +/{: x
> > assert. 0 < <./{. x NB. {.x is positive
> > 'd l'=.x
> > d <@]^:[&> (; l {.&.> 1) <;.1 y
> > )
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> ----------------------------------------------------------------------
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