Basically you are seeing the result of monadic =

l2

5 2 3 4 6 4 3 5 4 6 7 6

= l2

1 0 0 0 0 0 0 1 0 0 0 0

0 1 0 0 0 0 0 0 0 0 0 0

0 0 1 0 0 0 1 0 0 0 0 0

0 0 0 1 0 1 0 0 1 0 0 0

0 0 0 0 1 0 0 0 0 1 0 1

0 0 0 0 0 0 0 0 0 0 1 0

<@I. = l2

┌───┬─┬───┬─────┬──────┬──┐

│0 7│1│2 6│3 5 8│4 9 11│10│

└───┴─┴───┴─────┴──────┴──┘


I think you can simplify your Idot to:
   Idot=: [: <@I. =

It would be interesting to understand the actual problem you are trying to
solve with Idot.


On Mon, Jan 14, 2019 at 5:56 PM 'Skip Cave' via Programming <
[email protected]> wrote:

> Ok. You have convinced me to go with the empty box as a null indicator. In
> that case, we can make  the Idot verb dyadic, and generalize it:
>
>     Idot =.[:I.&.>[:{="1
>
> l1 =. 1 2 3 4 6 4 3 4 4 6 7 6
>
> 5 Idot l1
>
> ┌┐
>
> ││
>
> └┘
>
> l2 =. 5 2 3 4 6 4 3 5 4 6 7 6
>
> 5 Idot l2
>
> ┌───┐
>
> │0 7│
>
> └───┘
>
> ]m=.|:1 2 3 4,. 2 5 5 5,. 5 4 3 2 ,. 2 3 5 4,. 2 5 4 5
>
> 1 2 3 4
>
> 2 5 5 5
>
> 5 4 3 2
>
> 2 3 5 4
>
> 2 5 4 5
>
> 5 Idot m
>
> ┌┬─────┬─┬─┬───┐
>
> ││1 2 3│0│2│1 3│
>
> └┴─────┴─┴─┴───┘
>
> ]n=.|:2 5 5 5,. 1 2 3 4,. 5 4 3 2 ,. 2 3 5 4,. 2 5 4 5
>
> 2 5 5 5
>
> 1 2 3 4
>
> 5 4 3 2
>
> 2 3 5 4
>
> 2 5 4 5
>
> >m;n
>
> 2 5 5 5
>
> 1 2 3 4
>
> 5 4 3 2
>
> 2 3 5 4
>
> 2 5 4 5
>
> 1 2 3 4
>
> 2 5 5 5
>
> 5 4 3 2
>
> 2 3 5 4
>
> 2 5 4 5
>
>
> 5 Idot > m;n
>
> ┌─────┬─────┬─┬─┬───┐
>
> │1 2 3│ │0│2│1 3│
>
> ├─────┼─────┼─┼─┼───┤
>
> │ │1 2 3│0│2│1 3│
>
> └─────┴─────┴─┴─┴───┘
>
>
> NB. It's interesting what the monadic use of Idot does:
>
>
> Idot l1
>
> ┌─┬─┬───┬───────┬──────┬──┐
>
> │0│1│2 6│3 5 7 8│4 9 11│10│
>
> └─┴─┴───┴───────┴──────┴──┘
>
> Idot l2
>
> ┌───┬─┬───┬─────┬──────┬──┐
>
> │0 7│1│2 6│3 5 8│4 9 11│10│
>
> └───┴─┴───┴─────┴──────┴──┘
>
> Idot m ┌─┬─────┬─┬─┐ │0│1 │2│3│ ├─┼─────┼─┼─┤ │0│1 2 3│ │ │ ├─┼─────┼─┼─┤
> │0│1 │2│3│ ├─┼─────┼─┼─┤ │0│1 │2│3│ ├─┼─────┼─┼─┤ │0│1 3 │2│ │
> └─┴─────┴─┴─┘
>
> Idot n
>
> ┌─┬─────┬─┬─┐
>
> │0│1 2 3│ │ │
>
> ├─┼─────┼─┼─┤
>
> │0│1 │2│3│
>
> ├─┼─────┼─┼─┤
>
> │0│1 │2│3│
>
> ├─┼─────┼─┼─┤
>
> │0│1 │2│3│
>
> ├─┼─────┼─┼─┤
>
> │0│1 3 │2│ │
>
> └─┴─────┴─┴─┘
>
> Idot >m;n
>
> ┌─┬─────┬─┬─┐
>
> │0│1 │2│3│
>
> ├─┼─────┼─┼─┤
>
> │0│1 2 3│ │ │
>
> ├─┼─────┼─┼─┤
>
> │0│1 │2│3│
>
> ├─┼─────┼─┼─┤
>
> │0│1 │2│3│
>
> ├─┼─────┼─┼─┤
>
> │0│1 3 │2│ │
>
> └─┴─────┴─┴─┘
>
>
> ┌─┬─────┬─┬─┐
>
> │0│1 2 3│ │ │
>
> ├─┼─────┼─┼─┤
>
> │0│1 │2│3│
>
> ├─┼─────┼─┼─┤
>
> │0│1 │2│3│
>
> ├─┼─────┼─┼─┤
>
> │0│1 │2│3│
>
> ├─┼─────┼─┼─┤
>
> │0│1 3 │2│ │
>
> └─┴─────┴─┴─┘
>
> Can you explain what's going on here?
>
>
> Skip Cave
> Cave Consulting LLC
>
>
> On Sun, Jan 13, 2019 at 8:02 PM Ric Sherlock <[email protected]> wrote:
>
> > Depending on what you are trying to acheive, I think I'd represent the
> lack
> > of a match in a row as an empty rather than a _1:
> >
> > <@I."1 ] 5=m
> >
> > ┌┬─┬─┬─┐
> >
> > ││3│0│2│
> >
> > └┴─┴─┴─┘
> >
> >
> > Of course if you need the _1 then you can transform the above
> >
> > ;@([: (a:=])`((<_1),:~ ])} <@I."1) 5 = m
> >
> >
> > On Mon, Jan 14, 2019 at 10:42 AM 'Skip Cave' via Programming <
> > [email protected]> wrote:
> >
> > > What I would really like is for I. to return a _1 whenever there is no
> 1
> > in
> > > the match array, since there cannot be a negative index:
> > >
> > > I.5=1 2 3 4 6 4 3 4 4 6 7 6
> > >
> > > 4 8 10
> > >
> > > Idot 5=1 2 3 4 6 4 3 4 4 6 7 6
> > >
> > > _1
> > >
> > >
> > > ]m=.|:1 2 3 4,. 2 3 4 5,. 5 4 3 2 ,. 2 3 5 4
> > >
> > > 1 2 3 4
> > >
> > > 2 3 4 5
> > >
> > > 5 4 3 2
> > >
> > > 2 3 5 4
> > >
> > > 5=m
> > >
> > > 0 0 0 0
> > >
> > > 0 0 0 1
> > >
> > > 1 0 0 0
> > >
> > > 0 0 1 0
> > >
> > > ,Idot .5=m
> > >
> > > _1 3 0 2
> > >
> > >
> > > Can a verb Idot be designed, that does this?
> > >
> > > Skip Cave
> > > Cave Consulting LLC
> > >
> > >
> > > On Sun, Jan 13, 2019 at 2:41 PM Henry Rich <[email protected]>
> wrote:
> > >
> > > > Right.  Prefer (I.@:= ,) to I.@,@:= since it uses special code.
> > > >
> > > > Henry Rich
> > > >
> > > > On 1/13/2019 2:54 PM, 'Mike Day' via Programming wrote:
> > > > > You often see this sort of thing, returning pairs of indices of all
> > > > occurrences:
> > > > >
> > > > >     5 ($@] #.inv I.@,@:=) |: 1 2 3 4,. 2 3 4 5,. 5 4 3 2 ,. 2 3 5
> 4
> > > > > 1 3
> > > > > 2 0
> > > > > 3 2
> > > > >
> > > > > You can obviously get the row indices using {:”1 or some such, and
> > you
> > > > can of course make the bracketed code a named dyadic verb,
> > > > >
> > > > > Cheers,
> > > > >
> > > > > Mike
> > > > >
> > > > >
> > > > > Sent from my iPad
> > > > >
> > > > >> On 13 Jan 2019, at 17:55, 'Skip Cave' via Programming <
> > > > [email protected]> wrote:
> > > > >>
> > > > >> I know I can find the location (index) of a specific integer in a
> > > > vector of
> > > > >> integers using I.
> > > > >>
> > > > >> I.5=1 2 3 4 5 4 3 4 5 6 5 6
> > > > >>
> > > > >> 4 8 10
> > > > >>
> > > > >>
> > > > >> So I want to find the row index of a specific integer in an array
> of
> > > > >> integers:
> > > > >>
> > > > >> |:1 2 3 4,. 2 3 4 5,. 5 4 3 2 ,. 2 3 5 4
> > > > >>
> > > > >> 1 2 3 4
> > > > >>
> > > > >> 2 3 4 5
> > > > >>
> > > > >> 5 4 3 2
> > > > >>
> > > > >> 2 3 5 4
> > > > >>
> > > > >> 5=|:1 2 3 4,. 2 3 4 5,. 5 4 3 2 ,. 2 3 5 4
> > > > >>
> > > > >> 0 0 0 0
> > > > >>
> > > > >> 0 0 0 1
> > > > >>
> > > > >> 1 0 0 0
> > > > >>
> > > > >> 0 0 1 0
> > > > >>
> > > > >> ,I. 5=|:1 2 3 4,. 2 3 4 5,. 5 4 3 2 ,. 2 3 5 4
> > > > >>
> > > > >> 0 3 0 2
> > > > >>
> > > > >>
> > > > >> The first zero indicates that there is no 5 in the first row. The
> > > second
> > > > >> zero gives the index of the 5 in the third row. How can I tell
> > whether
> > > > the
> > > > >> zero is an index, or a null indicator?
> > > > >>
> > > > >> Skip
> > > > >>
> > ----------------------------------------------------------------------
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