Yes, it looks like we reached a similar conclusion. I just needed to step
back and think about / . Thanks

On Mon, Nov 12, 2018 at 7:34 PM Raul Miller <[email protected]> wrote:

> I didn’t see anyone using the approach which I would think of as obvious,
> so:
>
>    f=: fullNames
>
>    s=:shortNames
>
>    ,I.|:s +./@E.&>/ f
>
> 0 0 1 2
>
>
> Useful?
>
>
> Thanks,
>
>
> —
>
> Raul
> On Monday, November 12, 2018, Joe Bogner <[email protected]> wrote:
>
> > Thanks all. The solutions above were more complicated than I expected. I
> > had assumed I was missing something obvious. The obvious miss was to use
> /
> > ... I was thinking I could do something with rank alone
> >
> > For comparison it is combination of map and reduce in javascript
> >
> > ['Joe B','Joe Q','Bobby Z','Janet F'].map(function(x) { return
> > ['Joe','Bob','Jane'].reduce(function(acc,y,idx) { return
> > x.indexOf(y)>=0?idx:acc },-1) })
> >
> > [0, 0, 1, 2]
> >
> >
> >
> > On Mon, Nov 12, 2018 at 6:17 PM 'Mike Day' via Programming <
> > [email protected]> wrote:
> >
> > > or this variant?
> > >
> > >    f(1 i.~"1+./@E.~&> /)s. NB. f & s obvious?
> > > 0 0 1 2
> > >
> > > This lists the index of the first instance of a substring of each
> element
> > > of fullNames in the
> > > shortNames boxed list.  It assumes it's there somewhere, but you can
> use
> > > the property
> > > that non-inclusion is shown by index = length or number of shortNames.
> > >
> > > Any help,
> > >
> > > Mike
> > >
> > >
> > > > On 12 Nov 2018, at 22:29, Brian Schott <[email protected]>
> wrote:
> > > >
> > > > Is this anything you can use?
> > > >
> > > > I.fullNames(((]))&: >"0/-:"1 ((#@]{.[))& >"0/) shortNames
> > > >
> > > >
> > > > --
> > > > (B=) <-----my sig
> > > > Brian Schott
> > > >
> ----------------------------------------------------------------------
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> http://www.jsoftware.com/forums.htm
> > >
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