graphics - Character edge finding

The following line of code finds the edge of a character:

pic = Binarize[GradientFilter[Rasterize[Style["\[Euro]", FontFamily -> "Times"], 
                              ImageSize -> 200] // Image, 1]]

Mathematica graphics

The coordinates of the edges can be found as follows:

pdata = Position[ImageData[pic], 1];

Test:

Graphics[Point[pdata]]

Mathematica graphics

However, the points are not sorted in an order usable by Line or Polygon:

Graphics[Polygon[pdata]]

Mathematica graphics

This brings me to my question:

What would be an efficient method to sort the coordinates so that it would plot properly with Line or Polygon?

Additionally,

How to thin and smooth the set of points?

How to deal with characters with holes in them, like the ones below?

Mathematica graphics or

Answer

I think there is a neat solution. We have curios function ListCurvePathPlot:

pic = Thinning@Binarize[GradientFilter[Rasterize[Style["\[Euro]", 

FontFamily -> "Times"], ImageSize -> 200] // Image, 1]];

pdata = Position[ImageData[pic], 1];

lcp = ListCurvePathPlot[pdata]

enter image description here

Now this is of course Graphics containing Line with set of points

lcp[[1, 1, 3, 2]]

enter image description here

So of course we can do something like

Graphics3D[Table[{Orange, Opacity[.5],Polygon[(#~Join~{10 n})&
/@ lcp[[1, 1, 3, 2, 1]]]}, {n, 10}], Boxed -> False]

enter image description here

I think it works nicely with "8" and Polygon:

pic = Thinning@Binarize[GradientFilter[
Rasterize[Style["8", FontFamily -> "Times"], ImageSize -> 500] //Image, 1]]; 
pdata = Position[ImageData[pic], 1]; lcp = ListCurvePathPlot[pdata]

enter image description here

And you can do polygons 1-by-1 extraction:

Graphics3D[{{Orange, Thick, Polygon[(#~Join~{0}) & /@ lcp[[1, 1, 3, 2, 1]]]},
  {Red, Thick, Polygon[(#~Join~{1}) & /@ lcp[[1, 1, 3, 3, 1]]]},
  {Blue, Thick, Polygon[(#~Join~{200}) & /@ lcp[[1, 1, 3, 4, 1]]]}}]

enter image description here

=> To smooth the curve set ImageSize -> "larger number" in your pic = code.

=> To thin the curve to 1 pixel wide use Thinning:

 Row@{Thinning[#], Identity[#]} &@Binarize[GradientFilter[
 Rasterize[Style["\[Euro]", FontFamily -> "Times"], 
 ImageSize -> 200] // Image, 1]]

enter image description here

You can do curve extraction more efficiently with Mathematica. A simple example would be

text = First[
   First[ImportString[
     ExportString[
      Style["\[Euro] 9 M-8 ", Italic, FontSize -> 24, 

       FontFamily -> "Times"], "PDF"], "PDF", 
     "TextMode" -> "Outlines"]]];

Graphics[{EdgeForm[Black], FaceForm[], text}]

enter image description here

Comments

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Adding a thick curve to a regionplot

Suppose we have the following simple RegionPlot: f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}] Now I'm trying to change the curve defined by $y=g[x]$ into a thick black curve, while leaving all other boundaries in the plot unchanged. I've tried adding the region $y=g[x]$ and playing with the plotstyle, which didn't work, and I've tried BoundaryStyle, which changed all the boundaries in the plot. Now I'm kinda out of ideas... Any help would be appreciated! Answer With f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 You can use Epilog to add the thick line: RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}, PlotPoints -> 50, Epilog -> (Plot[g[x], {x, 0, 2}, PlotStyle -> {Black, Thick}][[1]]), PlotStyle -> {Directive[Yellow, Opacity[0.4]], Directive[Pink, Opacity[0.4]],

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