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Is it possible to identify a color (say grey in a 3 color Black, Gray and White image) and replace the color with HatchFilling?

I've tried to convert an image using ImageGraphics and then combine the result together inside of a Graphics with Hatchfill but I have had no luck. Is the result of ImageGraphics considered a Graphics expression?

labels = ClusteringComponents[img] Image[Replace[labels, ComponentMeasurements[{img, labels}, "MeanIntensity"], 2]]; MedianFilter[%, 1]; candidate1 = CurvatureFlowFilter[MeanShiftFilter[%, 8, 1/16], 2] (* Also have tried *) ColorQuantize[img, 4, Dithering -> False]; ImageApply[Max, ColorSeparate[%]] // ImageAdjust; smooth = CurvatureFlowFilter[MeanShiftFilter[%, 8, 1/14], 2]; MedianFilter[%, 1]; Blur[%, 2]

These are the current settings I am using to transform a picture into a smoothed out greyscaled image but trying Image

ImageGraphics[#, 3, Method -> {"LinearSeparable", 70 \[Degree]}, MinColorDistance -> .1, ImageSize -> 600] &@%

and then

graphics[{%,Hatchfilling}]

Does nothing. I know I need to throw in a colordetect for the color but I do not think that is the issue.

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    $\begingroup$ Most likely the answer is yes. Add some code to your question and there's a better chance for an answer that's useful to you. $\endgroup$ Commented Jan 23, 2021 at 19:00

2 Answers 2

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Using an input example from the documentation page for ImageGraphics:

hilbert = Import["https://i.sstatic.net/F9aZB.png"]

enter image description here

smoothHilbert = CurvatureFlowFilter[MeanShiftFilter[hilbert, 8, 1/16], 2]; ig = ImageGraphics[smoothHilbert, 3]

enter image description here

colors = Cases[ig, {a_, _FilledCurve} :> a, All]

enter image description here

ig /. colors[[1]] -> HatchFilling[]

enter image description here

ig /. {colors[[1]] -> HatchFilling[], colors[[2]] -> PatternFilling["Checkerboard", {10, 10}], colors[[3]] -> HatchFilling[-45 Degree, 5, 10]}

enter image description here

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  • $\begingroup$ awesome thank you so much for the help! $\endgroup$ Commented Jan 23, 2021 at 19:35
  • $\begingroup$ @skyfire, my pleasure. Thank you for the accept. $\endgroup$ Commented Jan 23, 2021 at 19:37
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    $\begingroup$ colors =Union[ Cases[ig, _?ColorQ, {1, -1}]] works fine too. $\endgroup$ Commented Jan 23, 2021 at 19:39
  • $\begingroup$ @andre314, thank you. It is much cleaner and seems more general. $\endgroup$ Commented Jan 23, 2021 at 19:48
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Here's a way to do it without converting the image into vectorized graphics.

hilbert = Import["https://i.sstatic.net/F9aZB.png"]; smoothHilbert = CurvatureFlowFilter[MeanShiftFilter[hilbert, 12, 1/16], 2]; quantizedHilbert = ColorQuantize[smoothHilbert, {Black, White, Gray}, Dithering -> False];

Mathematica graphics

whiteMask = ColorReplace[quantizedHilbert, Gray -> Black]

Mathematica graphics

{width, height} = ImageDimensions[whiteMask]; filling = Rasterize@Graphics[{ HatchFilling[], Rectangle[{0, 0}, {width, height}] }, ImageSize -> {width, height}, PlotRangePadding -> None]; filling whiteMask + ColorNegate[whiteMask] quantizedHilbert

Mathematica graphics

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