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11h
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11h
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1d
answered How to count proportion of two phase in a electron microscope picture
1d
comment Counting number of tubes in an image
Yes, but you want to change the x-coordinate in relation to the y-coordinate, and you want the singularity to be at y==vy, not y==0. So I think you need DataRange -> {{-vx, w - vx}, {0, h}-vy} (not tested)
1d
comment Counting number of tubes in an image
Your ImageTransformation isn't using vy at all? That can't be right...
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awarded  Enlightened
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Feb
9
awarded  Nice Answer
Feb
9
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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.