found it... seems the return values for 5!:1 are
character type and not numbers :(
{. will transform shape 1 into shape 0
(".({.,>{.(t=.>5!:1<'a'))) = 0 2 3 4 7
this runs into different problems when the result of
5!:1 is a verb.
--- p j <[EMAIL PROTECTED]> wrote:
> Stuck again, sorry.
> a=: 3 4
>
> (;{.(t=.>5!:1<'a'))
> 0
> t
> ┌─┬───┐
> │0│3 4│
> └─┴───┘
> $ (,,,.;{.(t=.>5!:1<'a'))
> 1
> (Can't seem to transform result of first expression
> into $ of 0 (blank result).. so that an expression
> like the following returns a 1 where it matches:
> (;{.(t=.>5!:1<'a')) (="0 1) 0 2 3 4 7
> 0 0 0 0 0
>
>
>
> --- "Miller, Raul D" <[EMAIL PROTECTED]> wrote:
>
> > Mark D. Niemiec
> > > Just because things look the same, doesn't mean
> > they are.
> >
> > But you could write a test that checks for "looks
> > the same"
> >
> > test=: -:&(((#~[:+./\1&~:)@$ ($,) ])@:":)
> >
> > Or, use test"0 if you prefer.
> >
> > Or, for more pedestrian uses, the ,. monad might
> be
> > sufficient
> > to coerce shapes.
> >
> > Of course, "looks the same" doesn't mean "behaves
> > the same"...
> > one=: 1 1 1 1 1 2$'1 '
> > one
> > 1
> > 1 + 1 2 3
> > 2 3 4
> > one + 1 2 3
> > domain error
> >
> > --
> > Raul
> >
>
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> >
>
>
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