I'm trying to build a real world map in Minecraft and want a pixel-perfect raster image I can copy pixel for block.

How can I modify

CountryData["World", {"Shape", "Robinson"}]

which gives

bad map

such that:

  • Each country is its own separate color (random colors are fine).
  • The water color is distinct from the background.
  • There are no black border lines (but the borders are still clear from the color difference in countries).
  • There is no anti-aliasing or color blending between regions of different colors, be they countries, water, or the background.
  • I can scale the map to the exact width (or height) in pixels I want.

Sorry if this is asking a bit much. Mathematica is completely new to me and I wasn't getting anywhere fumbling around with CountryData and GeoGraphics and WorldPlot for an hour.

  • $\begingroup$ What version are you on? $\endgroup$ Jul 22, 2016 at 5:02
  • 1
    $\begingroup$ @J.M. The pilot release for Raspberry Pi ( $\endgroup$ Jul 22, 2016 at 5:06
  • 2
    $\begingroup$ Would Graphics[Riffle[CountryData["World", {"SchematicPolygon", "Robinson"}] /. Polygon[{p__?MatrixQ}] :> Map[Polygon, {p}], Unevaluated[RandomColor[]], {1, -2, 2}], Background -> ColorData["Legacy", "Azure"]] suit your needs? $\endgroup$ Jul 22, 2016 at 5:20
  • $\begingroup$ @J.M. That still appears to have the anti-aliased borders between countries. I can maybe deal with those in an image editor but I'd prefer not to. $\endgroup$ Jul 22, 2016 at 5:29
  • $\begingroup$ What if you use "FullPolygon" instead of "SchematicPolygon"? FWIW: these are all generating vector graphics, which you can easily resize before saving as an image file. $\endgroup$ Jul 22, 2016 at 5:41

2 Answers 2


With GeoGraphics:

GeoGraphics[{GeoStyling@Opacity@1, RandomColor[], 
    CountryData[#, "SchematicPolygon"]} & /@ Join[CountryData["Continents"], CountryData[]], 
 GeoBackground -> Hue[0.56, .8, .8, .5], GeoRange -> "World", 
 GeoProjection -> "Robinson", Background -> White]

Mathematica graphics

  • $\begingroup$ Just what I wanted. (Any chance of adding antarctica?) $\endgroup$ Jul 22, 2016 at 6:59
  • 2
    $\begingroup$ Since Antarctica is not a country, it's not returned by CountryData[] (cf. this post). I've joined continents and countries to have everything - make sure you do not change the order as then continents would cover countries. $\endgroup$ Jul 22, 2016 at 7:49

From its doc

 2/3 Sqrt[
   1 - (CountryData[#, 
        "IndependenceYear"] /. {DateObject[{y_}] :> y, _Missing ->
CountryData[#, "SchematicPolygon"]} & /@ CountryData[]]

which makes it plot countries color coded by the length of their claimed independence. You can plot by any other method.

  • $\begingroup$ This is almost just what I want :) but (unless I'm very mistaken), it's not a Robinson projection. How could I fix that? $\endgroup$ Jul 22, 2016 at 5:28

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