Yes... I think... mostly.

J's native lists are densely packed lists, not linked lists.

Also, f should have been defined
   f=: 4:'0'

(A zero without the quotes means that the definition of f is a script
which appears on following lines and is terminated by a lone right
parenthesis.)

Thanks,

-- 
Raul

On Mon, Feb 22, 2021 at 5:00 PM Henry Rich <[email protected]> wrote:
>
> I am ignoring the 'linked' in 'linked list'.
>
> I think you are saying that f has natural rank 1 1: one row of x and one
> row of y.
>
> Define
>
> f =: 4 : 0
>
> And then use
>
> x f"1 1/ y
>
> / is equivalent to ("(lr,_))
>
> so this is
>
> x   f"1 1"1 _ y
>
> If I were you I would take whatever time it takes to see how that
> sentence operates.
>
> After you understand it, you could go back and write
>
> f =: 4 : 0"1 1    ( or f =: {{ ... }}"1 1 )
>
> and
>
> x f/ y
>
> Henry Rich
>
> On 2/22/2021 4:46 PM, Emir U wrote:
> > I'm trying to table a function run with various parameters against data. 
> > The function is in the form: x f y, where x are P parameters in a linked 
> > list, and y is the data which has N rows. f has to be explicitly defined 
> > because its complicated.
> >
> > Say X is Q different sets of parameters. I want to do:
> >
> > X f/ y
> >
> > and get a Q x N table where each cell represents the application of one 
> > parameter set to one data row.
> >
> > Even if a use:
> >
> > f=: 4 : 0 '0'
> >
> > which definitely returns a scalar, the table verb doesn't seem to think its 
> > rank 0 and won't behave as I'd like it to.
> >
> > I'd be grateful for some help on this.
> >
> > Emir
> >
> >
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>
>
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