# Tag Info

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PLEASE NOTE There is a reason why watermarks are there in the first place - to prevent unauthorised reuse of images. For more information on the issues around removing watermarks from images, this Wikimedia article is just one of many useful resources. The answer below is intended as an exercise in automatically removing text from an image ...

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An extended comment follows. Mondrian, in the late work referenced by the OP and characterized by primary colored rectangles separated by black lines, employed an extraordinarily sophisticated understanding of perception, color, and light. As background to understand what Mondrian does, I recommend The Interaction on Color, by Joseph Albers and Alfred C. ...

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Obtain the image: i = Import["http://i.stack.imgur.com/iab6u.png"]; Compute the distance transform: k = DistanceTransform[ColorNegate[i]] // ImageAdjust; ReliefPlot[Reverse@ImageData[k]] (* To illustrate *) Identify the "peaks," which must bound the Voronoi cells: l = ColorNegate[Binarize[ColorNegate[LaplacianGaussianFilter[k, 2] // ...

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This is a similar idea to ybeltukov's, iteratively splitting rectangles into two smaller rectangles. I've used an integer grid to avoid getting small offsets between adjacent blocks, and a weighted random choice to decide whether to split horizontally or vertically or do nothing. The idea is that long thin rectangles should preferentially be split along the ...

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TUTORIAL Import Image img = Import["https://i.stack.imgur.com/xzcUg.jpg"] Split into Components Using this approach (credit: nikie): m = MorphologicalComponents[Binarize@ColorNegate[ColorConvert[img, "Grayscale"]]]; Colorize[m] components = ComponentMeasurements[{m, img}, {"Area", "BoundingBox"}, #1 > 100 &]; trim = ImageTrim[img, #] & /@ ...

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Basically, you want to fit a shape to a set of points with outliers. One common algorithm to do this is RANSAC (random sample consensus). The basic outline of this algorithm is: Select N points at random (where N is the minimum number of points required for fitting the shape, i.e. 2 for a line, 3 for a circle and so on) Fit the shape to these points Repeat ...

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Here is my attempt to be Piet Mondrian colors = {RGBColor[0.9, 0.9, 0.9], RGBColor[0.05, 0.05, 0.05], RGBColor[0.8, 0.1, 0.1], RGBColor[0.1, 0.1, 0.5], RGBColor[0.9, 0.7, 0.1]}; split = # /. d : Rectangle[{x1_, y1_}, {x2_, y2_}] :> With[{t = RandomReal@BetaDistribution[10, 10], r = Random[]}, Which[ r < 0.3 (x2 - x1)/(y2 - y1), ...

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img = Import["ExampleData/lena.tif"]; Image[img, ImageSize -> 300] data = ImageData[img];(*get data*) {nRow, nCol, nChannel} = Dimensions[data]; d = data[[All, All, 2]]; d = d*(-1)^Table[i + j, {i, nRow}, {j, nCol}]; fw = Fourier[d, FourierParameters -> {1, 1}]; (*adjust for better viewing as needed*) fudgeFactor = 100; abs = fudgeFactor*Log[1 + Abs@...

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As was correctly noted in comments, out-of-focus images cannot been correctly detected by a simple gradient filter since out-of-focus images can have sharp edges. I propose another simple idea to detect such images. Introduction Roughly speaking, the brightness of a defocused image $B(x,y)$ is a convolution of a focused image $B_0(x',y')$ with some kernel $... 32 Without claiming much generality, I made the following. I'm using a slightly more complex image than your proposed one. i = Binarize@Import@"http://i.stack.imgur.com/qDby8.png"; idi = ImageDimensions@i; vertexI = SelectComponents[i, "Count", 5 < # < 100 &]; disk = 20 (*use some heuristics to ensure a ... 32 Another starting point, where the objects being more or less fixed size disks is used ad hoc to measure their centroids as components after some mangling, and those which are close enough to each other are connected in the graph if out of sample of four hundred points along the edge at most four are not "white." Image is in the variable img and plenty of ... 30 I think the essence of the problem here is that width needs to be counted orthogonally to some best fit line going through the elongated shape. Even naked eye would estimate some non-zero slope. We need to make line completely horizontal on average. We could use ImageLines (see this compact example) but I suggest optimization approach. Import image: i = ... 29 Version 9 answer - use built-in functionality The symbols ImageCreateHDRI and ImageToneMapHDRI were present in version 8 but didn't seem to do anything. In version 9 there is functioning code behind them. This is all undocumented, and therefore liable to change before it is officially released, but here is what I've managed to dig up. Image`CreateHDRI ... 29 This version of generation code is originally posted at wolfram community. I guess a story-telling type post would attract more upvotes and probably give some insight about how to 'solve problems' using Mathematica, so I would go into details and try to explain not only the code but also how I figured out how to write them. How this works? In fact, this ... 28 Preload all chemical data: ChemicalData[All, "Preload"]; RebuildPacletData[]; (* the latter should not really be necessary *) Get all names: cd = ChemicalData[]; Get their molecular formulae: l = ChemicalData[#, "MolecularFormulaString"] & /@ cd; By counting the Cs, Os and Hs in the tattooed diagram we know we have to find$\rm{C_{19}H_{28}O_{2}}\$. ...

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i = Import@"http://i.stack.imgur.com/8I3B1.jpg"; f[{{tmin_, tmax_}, {rmin_, rmax_}}, ___] := Module[{l = Join[{{0, 0}}, Table[{Cos@t, Sin@t}, {t, tmin, tmax, (tmax-tmin)/100}]]}, {Texture[i], EdgeForm[], Polygon[l, VertexTextureCoordinates -> 1/2 Transpose[Transpose[l] + {1, 1}]]}] Framed@PieChart[{1, 2, 3, 4, 5, 6}, ChartElementFunction -> f] ...

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Here a quick hack for PNG images. As its Wikipedia page shows the format works with coded chunks and you can make up and insert chunk types yourself. I'm not sure how safe it is to add beyond the official end of file marker as Simon Woods suggests in his answer. It seems like a breach of the standard to me. The following code, which more closely seems to ...

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Update 2: The function projectToWalls does not work in version 12.0 because the function PlotRange no longer works. To fix the issue, replace PlotRange with plotRange where plotRange = PlotRange/.AbsoluteOptions[#, PlotRange] &; in the definition of projectToWalls. Original answer: You can post-process a Graphics3D object to project the lines to the ...

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One standard way to detect circular shapes is to binarize the image and apply a distance transform: The maxima locations of the distance transform are the centers of the circles. To make this work on your ellipses, I first have to stretch them to be (roughly) circular, as @Rahul Narain suggested in a comment: img = ColorConvert[Import["http://i.imgur.com/...

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No, they are not copyright-free, but the data they are based on is openly licensed. The map data mainly comes from OpenStreetMaps (which is licensed under the Open Database License, similar to Creative Commons) and so you will need to include OpenStreetMaps Contributors in your attribution. If you hover over a GeoGraphics, you can see Wolfram's attribution ...

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Tricky. But with a bit of creative cheating, I can get close: First, load the image and binarize it: img = Import["http://i.imgur.com/qAZBdFb.jpg"]; bin = MorphologicalBinarize[ColorNegate@img, {.1, .5}] I invert the image for two reasons: First, MorphologicalBinarize takes a lower and an upper threshold, i.e. it assumes bright blobs on dark ...

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One simple way of "filtering" your data is to treat the points as a graph, and search for the shortest path from left to right: xScale = 10.; xy = Transpose[{N[Range[Length[data]]]*(xScale/Length[data]), data}]; start = {-xScale, Mean[data]}; finish = {2*xScale, Mean[data]}; graph = NearestNeighborGraph[Join[{start}, xy, {finish}], 25]; graph = ...

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When thinking about graphics formats that can be displayed in web browsers and also in Word, the first thing that comes to mind is a rasterized image. However, there is one alternative that makes including comments a complete no-brainer: SVG (scalable vector graphics). The way you do it is similar to what cormullion suggested for EPS, except that EPS of ...

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A few minor mistakes: corners and the target coordinates weren't in the same order For some reasons, pixels isn't the default unit for ImagePerspectiveTransformation and friends - you have to specify PlotRange and DataRange explicitly the target coordinates should go first in FindGeometricTransform - alternatively, you can pass InverseFunction@ft[[2]] to ...

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Reading the file as Import["file.jpeg", "ImageNoExif"] is noticeably faster on my machine than the default Import["file.jpeg", "Image"] Benchmark in 11.1: Import["~/Desktop/Untitled.jpg", "Image"]; // RepeatedTiming (* {0.054, Null} *) Import["~/Desktop/Untitled.jpg", "ImageNoExif"]; // RepeatedTiming (* {0.025, Null} *) This is not a full solution, ...

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Here's a method based on creating a MeshRegion from the text: text = Style[HoldForm @ Sum[x^2, {x, 0, 10}], 100, Bold]; graphics = First[text ~ExportString~ "PDF" ~ImportString~ "PDF"]; region = DiscretizeGraphics[graphics, MaxCellMeasure -> 5]; image = ExampleData[{"ColorTexture", "Kingwood"}]; RegionPlot[region, Frame -> False, ...

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i = Binarize@Import["http://i.stack.imgur.com/qofeF.png"]; vertexI = SelectComponents[i, "Count", 10 < # < 100 &]; vxPos = ComponentMeasurements[vertexI, "Centroid"]; lines = Subsets[Range@Length@vxPos, {2}]; linePos = lines /. vxPos; ti@x_ := Total@Flatten@ImageData@Binarize@x p = Position[ti@i - ti@Show[i, Graphics@Line@#] & /@ ...

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Here is a Nearest-based method. This is quite similar to what @Mr. Wizard did for approximating 3D (ordinary) Voronoi. comps = MorphologicalComponents[img]; cmap = Flatten[MapIndexed[#2 -> #1 &, comps, {2}]]; comparray = DeleteCases[cmap, _ -> 0]; nf = Nearest[comparray]; Now we build the table giving Voronoi components. Timing[ ...

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Too late and not so general as @Nikie but I will still like to answer in order to highlight a method involving generalized Eigen values! Lets take the image. img = Import["http://i.stack.imgur.com/H63BK.jpg"]; (* Detecting ellipsoidal edges and removing the straight line component *) i = ImageSubtract @@ {EdgeDetect[img, 10],EdgeDetect[img, 10, "...

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