Skip to main content

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 Mathematica graphics



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

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...