# Tag Info

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Result first, I hope that would be good enough: The idea is quite simple------Find those feature and try to delete them. How to use the code? Manually select the features you would like to delete, binarize it properly and put it in ker. For example, at this place you could use something like this: If you want a better result, create another ...

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A function to subdivide the triangles based on their gray level: h[{v1_, v2_, v3_}] := With[{a = EuclideanDistance[v1, v2], b = EuclideanDistance[v1, v3], c = EuclideanDistance[v2, v3]}, With[{s = (a + b + c)/2}, (2 Sqrt[s (s - a) (s - b) (s - c)])/c ]] shadeTri[tri_, col_, f1_: 1, fc_: 1] := If[col > .8, tri, With[{v = tri[[1, ...

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One possibility would be to apply a highpass filter, since what you want to remove is the slowly undulating background. img = Import["http://i.stack.imgur.com/rvRAc.png"]; imgHP = ImageAdjust[HighpassFilter[img, 0.05]] then binarize: Binarize[imgHP] To follow your original idea of locating the background and subtracting it -- this can be ...

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(These solutions loosely adhere to OP's request in the question. They were mostly made because of similar art/solutions pointed in the comments.) "Delaunay raster" like Here is a solution related to "Delaunay raster" discussed in the question comments. It is based on the Mathematica documentation page "Create a Mesh Region from Image Data". I changed ...

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Here's something to get the ball rolling: img = ColorConvert[Import["http://i.stack.imgur.com/FaE06.jpg"], "Grayscale"]; DelaunayMesh[ ImageCorners[img, 1, 70*^-6], MeshCellStyle -> {{2, All} -> White, {1, All} -> GrayLevel[0.5], {0, All} -> Black} ] I'm looking forward to better results :-)

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Where can the code be improved? First of all: Try to avoid loops (Do, For, While) where possible - they're usually much more verbose than functional alternatives. Second: Read the documentation. ComponentMeasurements can take a list of measurements, and will then return a list of values for each component. Third: Use functions to encapsulate units of ...

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img = Import["http://i.stack.imgur.com/bkNTj.png"]; pos = ComponentMeasurements[img// Binarize, {"Centroid", "EquivalentDiskRadius"}][[All, 2, 1]]; Length[pos] Show[img, Graphics[{Red, Circle[#, 10]} & /@ pos]] 3 You might be interested in Count flowers in an image and Count Elements in Image. For your complete folder, SetDirectory[...

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Prologue Some five years ago I have asked exactly this question to the Wolfram support people. Below I have taken their respective answers (one sentence for each Method) but have added a lot of further reading. Finally, in an Add-On I demonstrate my own implementation of Rosenfeld's 1971 variant of a 2D thinning algorithm, in order to let you compare a few ...

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This seems pretty fast (it takes roughly 0.08s on your set of images): images = Normal@Databin["dsNFr2og"]; Total[#, 2] & /@ ImageData /@ Binarize /@ images; % / (Times @@ ImageDimensions[images[[1]]]) // N; Extract[images, Position[%, p_ /; p < 0.5]] Adjust the 0.5 threshold to your liking to retain more or fewer images. (Thanks nikie for the ...

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Yesterday I showed how to locate a subimage inside a large image using ImageCorrelate. I will now show how to use the same approach to solve this problem. Start by importing the images that you want to compose: i1 = Import["http://i.stack.imgur.com/UjGAz.jpg"]; i2 = Import["http://i.stack.imgur.com/93kwm.jpg"]; GraphicsRow[{i1, i2}] Since the right ...

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Here's my approach which returns each color component's percentage in a given bin. First, some preprocessing. This quantizes the colors using the Nearest function because by default Mathematica's ColorQuantize fails to properly quantize the image. i = Import["http://i.stack.imgur.com/otAao.jpg"]; ncolors = 4; dc = DominantColors[i, ncolors]; dcl = dc /. ...

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This is not very automated but perhaps it will motivate experts: Copied the images Changing image to graphic and finding corresponding points: fun[img_] := With[{id = ImageDimensions[img]}, {Texture[img], Polygon[{{1, 1}, {id[[1]], 1}, id, {1, id[[2]]}}, VertexTextureCoordinates -> {{0, 0}, {1, 0}, {1, 1}, {0, 1}}]}] {ma, mb} = ...

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There's a few steps in solving this problem: Separate colors completely Partition the image Compute the correlation between neighbouring picture segments. To solve the first problem, I use ColorQuantize first in order to make similar colors the same to avoid noise. Then by using LinearSolve, I can change the three colors to R/G/B so that I can use ...

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I am going to copy-paste Simon Wood's answer of How to create a new “person curve”? param[x_, m_, t_] := Module[{f, n = Length[x], nf}, f = Chop[Fourier[x]][[;; Ceiling[Length[x]/2]]]; nf = Length[f]; Total[Rationalize[2 Abs[f]/Sqrt[n] Sin[Pi/2 - Arg[f] + 2. Pi Range[0, nf - 1] t], .01][[;; Min[m, nf]]]]] tocurve[Line[data_], m_, t_] := ...

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The most straightforward approach is ImageCorrelate. I'll show you an example. nikie wrote an excellent answer explaining this method here. large = ExampleData[{"TestImage", "Boat"}] smaller = ColorConvert[Import["http://i.stack.imgur.com/NAHqc.png"], "Grayscale"] corr = ImageCorrelate[large, smaller, EuclideanDistance]; ImageAdjust[corr] The ...

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Here is what I get. img = Import["http://i.stack.imgur.com/o3S3Q.jpg"]; red = Fold[ImageSubtract, #[[1]], Rest[#]] &@ColorSeparate[img] b = Binarize[red] pos = ComponentMeasurements[b, {"Centroid", "EquivalentDiskRadius"}]; Length[pos] 633 Fine Tuning Since the flowers in the top left corner are comparatively dull in colour, they are ...

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