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| bio | website | |
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| location | ||
| age | ||
| visits | member for | 1 year, 3 months |
| seen | 52 mins ago | |
| stats | profile views | 191 |
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Dec 26 |
comment |
Grid without lines, but with dots You could replace the extra dots with spaces, by inserting this between Transpose and Grid: MapIndexed[If[OddQ@Total@#2, " ", #1] &, #, {2}] & |
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Dec 23 |
comment |
Dynamic Graph visualization Having looked through the options in the GraphPlot documentation, my guess is that what you want is only possible if you write your own force-directed graph layout routine and animate the vertices yourself. In my comments below I was just trying to suggest how to get as close as possible to a nice result with only the existing GraphPlot functionality. |
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Dec 23 |
awarded | Critic |
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Dec 23 |
comment |
Time-series decomposition in Mathematica Well, I went ahead and found the documentation: stat.ethz.ch/R-manual/R-devel/library/stats/html/decompose.html. It seems pretty easy to implement: just a bunch of moving averages. The period of seasonality is assumed to be 1 unit (12 data points in this case), so no need for a heavy-duty NonlinearModelFit. |
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Dec 23 |
comment |
Time-series decomposition in Mathematica This question would be easier to answer if you could find a reference to the algorithm that R's time series decomposition is using. It might even be described in the documentation. Asking to reverse-engineer the functionality of a magical third-party black box is still a valid question, but a much harder one (as evidenced by your comment on b.gatessucks's answer). |
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Dec 23 |
comment |
Dynamic Graph visualization @Sean: Perhaps I wasn't clear. What I meant by "all the input edges enabled regardless of lifetime" was that you do a GraphPlot[#1 -> #2 & @@@ dynamicGraph] and then keep those vertex positions fixed while a varying subset of edges is displayed. |
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Dec 23 |
comment |
Dynamic Graph visualization @Sean, RiemannZeta: What if you set the precomputed vertex placement by doing a GraphPlot with all the input edges enabled regardless of lifetime? Then, granted the layout won't adjust itself as the structure changes over time, but it would still reflect the "average" structure of the graph to some extent. |
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Dec 23 |
comment |
How do I make an image “nearly” black and white? If you replace the Blur with an Image[#, ImageType[im1]]&, you get smooth contours without the extra blurring: i.stack.imgur.com/dI9XW.png. The only reason it didn't work without Blur was that Binarize implicitly converted the ImageType to Bit. P.S. You could also use Scaled[4] instead of 2500 if you want it to work without modification for images of different sizes. |
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Dec 21 |
awarded | Good Answer |
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Dec 21 |
comment |
Creating an animation illustrating the time-evolution of a pre-computed orbit By the way, this is not really simulation; this is just animation. Simulation would be if you gave us only the inverse square law and the initial state of the particle, and asked us to plot its orbit. |
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Dec 20 |
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FindFaces: how to improve the results of the face recognition feature Mr. Angry Physicist never goes away though. I think he looks like Freud. |
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Dec 20 |
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FindFaces: how to improve the results of the face recognition feature The false positives appear to be very sensitive to denoising. If you run, say, a TotalVariationFilter on the image, many of the false positives go away and new false positive appear (the new ones are all different for Method -> "Gaussian", "Laplacian", or "Poisson"). The obvious solution: apply a bunch of different filters to the image, do FindFaces on the results, and keep the faces that appear in all of them. :) |
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Dec 20 |
comment |
Drawing a torus not bounding a solid torus From the documentation: " Plot has attribute HoldAll and evaluates $f$ only after assigning specific numerical values to $x$.
In some cases, it may be more efficient to use Evaluate to evaluate $f$ symbolically before specific numerical values are assigned to $x$." The same is true of ParametricPlot3D and most other plotting functions. |
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Dec 20 |
answered | Drawing a torus not bounding a solid torus |
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Dec 20 |
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Drawing a torus not bounding a solid torus But your surface does go through the origin! The knot goes through $(1,0,0)$ instead. |
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Dec 19 |
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Drawing a torus not bounding a solid torus Your knot surface has self-intersections, which make it look very confusing when you turn it inside out. You could try making it less thick. Beyond that, I'm with wxffles here: it would really help if you could explain more clearly what you mean by "drawing this torus the way I want". Perhaps with some sort of a sketch of the shape you expect to see? Anyway, I played around with it and the best I could get is i.stack.imgur.com/sPXs0.png |
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Dec 19 |
awarded | Nice Question |
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Dec 16 |
awarded | Mortarboard |
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Dec 16 |
awarded | Nice Answer |
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Dec 16 |
revised |
How to generate a picture like this? avoid the distracting horizontal lines by using a different base mesh |
