Devon:
As a start, I suggest binning the points in equal angled wedges and then
selecting the furthest 2 points in each wedge.
ppbin=:<.10* {:"1 pp
top2=:2 {. \:~
perim=:;top2&.> ppbin < /. pp
perimc=:p2c"1 perim
viewmat (2) (<"1]1+perimc-"1 <./pts)}1 bordFill pts
You probably really need thinner near-horizontal wedges than
near-vertical, but the viewmat picture looks more like what I think you
are after.
Best wishes,
John
Devon McCormick wrote:
> Members of the Forum -
>
> I'm stymied by what seems like a simple problem that goes something like
> this: I have a set of points longer in one direction than the other, like
> this:
>
> $pts=. <.(]-"1 mean) ~.|:+/&>?&.>(<3 1000)$&.>20 80 NB. mean=: +/%#
> 913 2
> viewmat 1 bordFill pts
> NB. where
> bordFill=: 4 : '(1)(<"1 y-"1 x-~<./y)}0$~2$(>:+:x)+(>./y)-<./y' NB. x
> border around integer point pairs y
>
> I want to identify points near the perimeter. One idea I had for doing
> this
> is to use polar coordinates,
>
> pp=. c2p"1 pts NB. Polar points: length, angle
> NB. Where:
> p2c=: +. @ (r./) NB.* p2c: polar to cartesian (point
> pair)
> c2p=: ({.,2p1&|@{:)@:(*. @ (j./)) NB.* c2p: cartesian to polar (point
> pair)
>
> I want to get a rough idea of the perimeter by dividing the angles into,
> say, 32 even groups and selecting the 10 most distant points within each
> "wedge".
>
> gv=. /:1{"1 pp NB. Order by angle
>
> I'm not sure why the following does not seem to work correctly:
>
> lens=. (<.32*gv%#gv) </. 0{"1 pp NB. Group lengths by angle
> sector
> farpps=. (<.32*gv%#gv) </. pts NB. Same grouping for points
> $farpps=. ;((10{.\:)&.>lens){&.>farpps NB. Points corresponding to 10
> longest lengths
> 320 2
>
> [The expression "<.32*gv%#gv" is supposed to map the grade vector to key
> values 0-31.]
>
> If this worked correctly, I'd expect the following image to show a ring of
> differently-colored points as a perimeter:
>
> viewmat (2) (<"1]1+farpps-"1 <./pts)}1 bordFill pts
>
> Instead, the coloring seems to apply only to the points furthest from the
> center, which tend to be those on either side along the longer axis.
>
> Does this make sense? Can anyone see what I'm doing wrong or suggest
> another method?
>
> Thanks,
>
> Devon
> --
> Devon McCormick, CFA
> ^me^ at acm.
> org is my
> preferred e-mail
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