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Author Comment/Response
06/29/12 7:34pm

In Response To 'Re: Re: PointSize'
Here's a better way: Find a bounding polygon (a rectangle); reverse the inside polygon (switches inside and outside); stitch together with bounding polygon.

A little tricky because a country's polygon is really a list of polygons. I didn't think it would work because of the edges that connect the polygons, but Mathematica treats them as zero width; so it works! I hope the code is not mysterious.

Stitch[inPolys_, outPoly_] :=
Fold[Function[{out, in},
Join[out, {out[[1]], in[[1]]}, Reverse[in]]
], outPoly, inPolys];
Module[{ctry, bBox, padding, plRange},
ctry = CountryData["Spain", "Polygon"][[1]];
padding = 0.25;(* to make bBox larger *)

plRange = ({Min[#1], Max[#1]} &) /@ Transpose[Join @@ ctry];
bBox = Tuples[
plRange + {{-padding, padding}, {-padding, padding}}][[{1, 3, 4,
Graphics[{(*Coordinate of the point at {-3.71`,40.42`}.*)
Red, {PointSize[pointSize],
40.42`}]},(*Border of country whit polygon*)
0.9, 0.9], Polygon@Stitch[ctry, bBox]}},
Background -> RGBColor[0.4, 0.4, 0.4], PlotRange -> plRange,
PlotRangePadding -> 0.2, PlotRangeClipping -> True]], {pointSize,
0, 1}]

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Subject (listing for 'PointSize')
Author Date Posted
PointSize MaxiWenz 06/25/12 09:52am
Re: PointSize Michael 06/25/12 6:05pm
Re: Re: PointSize MaxiWenz 06/28/12 03:36am
Re: Re: PointSize MaxiWenz 06/28/12 03:57am
Re: Re: Re: PointSize Michael 06/29/12 7:34pm
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