Nice! Xiao-Yong's verb does exactly what I need.
f=:(] +:@{`[`]} <@<@<@]{[)"_ 0 [:I.2=/\]
t =. 1 2 2 4 1 5 3 4 4 4 2 3 3
f t
1 4 4 1 5 3 4 4 4 2 3 3
1 2 2 4 1 5 3 8 4 2 3 3
1 2 2 4 1 5 3 4 8 2 3 3
1 2 2 4 1 5 3 4 4 4 2 6
I can understand [:I.2=/\]. Using I. is the perfect way to find the
indices of the dups. However, the rest of that big infinite-rank tacit
expression to the left must generate each of the reduced items. I'm totally
lost trying to figure that out. Why the zero? Why three <@? I thought [
represented a left argument, but there isn't a left argument. Any help is
appreciated.
Skip
Skip Cave
Cave Consulting LLC
On Mon, Nov 7, 2016 at 10:25 PM, Xiao-Yong Jin <[email protected]>
wrote:
> So you probably just need this:
> f=:(] +:@{`[`]} <@<@<@]{[)"_ 0 [:I.2=/\]
> or really the adverb
> a=:1 :'u@(] +:@{`[`]} <@<@<@]{[)"_ 0 [:I.2=/\]'
>
> You may find different solutions if you abstract your problem differently.
>
> > On Nov 7, 2016, at 9:30 PM, Skip Cave <[email protected]> wrote:
> >
> > Very interesting. Quite a difference in space and time usage, not to
> > mention length of the tacit code.
> >
> > For my application, a rank 0 or 1 in the summed result isn't all that
> > critical. As long as the contents of each box can be used in further
> > arithmetic and Boolean operations (e.g. it's not text), then I'm ok.
> >
> >
> > It looks like nouns with ranks 0 or 1 don't affect the outcome of simple
> > arithmetic operations:
> >
> >
> > ($4);($,4);(4=,4);(1+4);(1+,4);((1+4)=1+,4);($1+4);($1+,4)
> >
> > ┌┬─┬─┬─┬─┬─┬┬─┐
> >
> > ││1│1│5│5│1││1│
> >
> > └┴─┴─┴─┴─┴─┴┴─┘
> >
> > The only time the difference becomes an issue:
> >
> > ($4)=($,4)
> >
> > |length error
> >
> > | ($4) =($,4)
> > But I don't care about this.
> >
> > Actually, Mike Day spotted what I was really trying to attempt.
> > The original starting vector is:
> >
> > n =: 1 2 2 4 1 5 3 4 4 4 2 3 3
> >
> > Then I box at each possible pair of integers using an infix
> sliding-window
> > on each pair of integers. This sliding box scheme was just to identify
> all
> > the duped pairs in the original vector.
> >
> > ]N =: 2<\pp
> >
> > ┌───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┐
> >
> > │1 2│2 2│2 4│4 1│1 5│5 3│3 4│4 4│4 4│4 2│2 3│3 3│
> >
> > └───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┴───┘
> >
> >
> > What I ultimately wanted is a verb that will replace the duped pairs in
> the
> > *original* un-boxed vector, one at a time:
> >
> > n =: 1 2 2 4 1 5 3 4 4 4 2 3 3
> >
> > f n
> >
> > 1 4 4 1 5 3 4 4 4 2 3 3
> >
> > 1 2 2 4 1 5 3 8 4 2 3 3
> >
> > 1 2 2 4 1 5 3 4 8 2 3 3
> >
> > 1 2 2 4 1 5 3 4 4 4 2 6
> >
> > Of course, the final row lengths will be one shorter that the original
> row
> > length.
> >
> > Skip
> > ----------------------------------------------------------------------
> > For information about J forums see http://www.jsoftware.com/forums.htm
>
> ----------------------------------------------------------------------
> For information about J forums see http://www.jsoftware.com/forums.htm
>
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
