nikie
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 Feb 8 answered Boundary identification Feb 6 answered Is there a bug of InverseDistanceTransform? Feb 6 comment Is there a bug of InverseDistanceTransform? Try e.g. InverseDistanceTransform[ImageMultiply[pic, 10]] to get disks with radius 10 Feb 6 comment Is there a bug of InverseDistanceTransform? If you evaluate Max[ImageData[pic]] with the image in the documentation, you get 13, i.e. the values of the white pixels are larger than 1, namely the disk radii in pixels. I'm guessing your image contains values between 0 and 1, so InverseDistanceTransform creates disks with radii between 0 and 1 Feb 4 answered Follow each pixel brightness in set of images (Memory constrained) Feb 3 comment Improving visibility of text in scanned image Related: dsp.stackexchange.com/a/1934/291 Feb 1 comment NelderMeadMinimize for 2D Gaussian fitting Dumb question: why not just take the log of the images, then fit a 2nd order polynomial to them? That way, it's a linear least squares problem, and you can use LeastSquares Feb 1 answered extract data from a graph image Jan 31 comment Can I get the curvature at any point of a random curve? @yode: You mean instead of the spanning tree? But the shortest tour (or any tour) goes through all points, then returns to the beginning. I don't want to visit all points, I want to remove those little "branches" that Thinning produces. And I don't want my path the return to the beginning. Also, finding a the shortest tour is an NP-complete problem where only approximate solutions are realistically possible, while finding a spanning tree is cheap. Jan 31 answered Counting number of tubes in an image Jan 30 comment Grayscale Morphology I'm just translating the formulas on the wikipedia page azt linked. Apparently, to get the same result as in binary morphology, the kernel should contain 0 and -infinity: greyKernel = DiskMatrix[4] /. {0 -> -\[Infinity], 1 -> 0}; fil = ImageFilter[Max[# + greyKernel] &, img, 4] Jan 29 comment Grayscale Morphology I think (?) it should be ImageFilter[Max[# + ker] & ... Jan 29 awarded Good Answer Jan 29 comment Convex and concave regions of a closed curve Jan 26 awarded Yearling Jan 26 comment Convert (rescale) 16 Bit png image Why don't you just use imageData/8? Jan 25 comment Get an array of coordinates of the edges of an image Note that Position[ImageData[...]] returns indices, not coordinates. (The distinction is a bit confusing in MMA image processing, see for example: mathematica.stackexchange.com/a/104047/242). Use PixelValuePositions[edge, 1] to get the coordinates. Rule of thumb: If you want to manipulate the raw array returned by ImageData, use indices. If you want to draw something over the image using e.g. Show, use coordinates Jan 24 comment How to get a full closed contour image In coordinatescontour = [...] // DeleteSmallComponents the DeleteSmallComponents is probably a mistake. Other than that, your code seems to do what you want already. Jan 20 answered Fitting a curve to an image object: not a superellipse, not an egg curve.. what else? Jan 15 revised Getting image data values with coordinates in circular region added 33 characters in body