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location Munich, Germany
age 36
visits member for 2 years, 7 months
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I work at a small mechanical engineering company, where I develop software and image processing algorithms for camera-based inspection machines.


Aug
3
answered Computing a mean of median list of points
Aug
1
revised Generating an “average curve” from a dense set of “semicontinuous curves” which clear trends
added 436 characters in body
Aug
1
comment Generating an “average curve” from a dense set of “semicontinuous curves” which clear trends
I'm not quite sure I know what you meant, so I've updated the answer, and it seems to work fine.
Aug
1
revised Generating an “average curve” from a dense set of “semicontinuous curves” which clear trends
added 1463 characters in body
Aug
1
comment Generating an “average curve” from a dense set of “semicontinuous curves” which clear trends
Ah, I'm sorry, I didn't read your comment properly. I thought the data was in format {{y11,y12,y13},{y21,y22},{},{y41,y42,y43}..., i.e. a list for each x-value which contains only the y-values at that location. Then you could have added Flatten to makeHist and everything else should have worked automatically. Can you maybe transform your data to that format?
Aug
1
comment Generating an “average curve” from a dense set of “semicontinuous curves” which clear trends
@SparsePine: if it's always the first element, you could maybe index the set to drop that element data[[All,All,2;;]]?
Aug
1
comment Generating an “average curve” from a dense set of “semicontinuous curves” which clear trends
@SparsePine: Without having tested this: I think all you have to do is adjust the makeHist function so it Flattens the data parameter before passing it to BinCounts
Aug
1
answered Generating an “average curve” from a dense set of “semicontinuous curves” which clear trends
Jul
23
awarded  Nice Answer
Jul
17
awarded  Nice Answer
Jul
1
comment Is there an easy way to create a “IntensitySquaredCentroid” using ComponentMeasurements?
Why not just pass Image[ImageData[image]^3] to ComponentMeasurements?
Jun
27
comment Circle as a function parameter
If it's always 3 circles, you can just write findCircle[Circle[c1_,r1_],Circle[c2_,r2_],Circle[c3_,r3_]] := and use the centers c1..c3 and radii r1..r3 in the function body
Jun
27
comment Crash when displaying / rasterizing graphics with textures at high resolutions
Weird. On my Windows machine, I can draw with ImageSize->10000 and highest graphics quality, no problem. Maybe anti-aliased drawing is done by the OS and OSX can't draw to surfaces bigger than the screen resolution?
Jun
18
comment How could I define a function as the solution of equations in Mathematica?
Isn't it inefficient to call Solve every time the function a is evaluated? If you used Set instead of SetDelayed, the equation would only be solved once.
Jun
14
awarded  Nice Answer
Jun
14
comment Digital filter of image in Fourier space
@Andrei: It's weird: if I import the data from Pastebin, and look at the fourier transform, it still doesn't look nearly as smooth and artifact-free as the one in your question.
Jun
14
comment Find radii of concentric circles in image
The coordinate conversion was off by one. Fixed it. I love MMA for image processing, but the distinction between indices and coordinates is ugly. Especially as there is no built-in function to convert them.
Jun
14
revised Find radii of concentric circles in image
Off-by-one error
Jun
14
comment Find radii of concentric circles in image
@cormullion: I'm not 100% sure. Since the center is a least squares estimate using all gradients in the image, the error could be due to gradients that don't point towards the center. Or maybe my index -> coordinate conversion is off by one?
Jun
14
answered Find radii of concentric circles in image