plotting - Why is ListPlot so slow here?

While examining How can I monitor the progress of a Plot? I was surprised to discover that in some cases ListPlot in version 10.0 and 10.1 is orders of magnitude slower than it is in version 7. This is not rendering time but generation of the Graphics itself. Here is an example.

dat = Table[{x, y}, {x, 200}, {y, RandomReal[9, 500]}];

lp = ListPlot[dat, ImageSize -> 600]

enter image description here

Rendering this plot takes only ~0.08 second according to EvaluationCompletionAction -> "ShowTiming" as seen by evaluating lp separately. However generating lp (in 10.1) is quite slow:

ListPlot[dat] // RepeatedTiming // First

2.02

This takes only 0.018 second in version 7. Why is 10.1 two orders of magnitude slower?

David Skulsky reports these AbsoluteTiming results:

MacBook Air: v8 2.1 sec, v9 0.43 sec, v10 3.6 sec.

Apparently the problem is not limited to v10 though it is most severe there. Should this not be a simple operation and much faster than this as indeed it was in version 7?

First attempt at analysis

Since no useful explanation had yet been provided I thought I would see if I could learn anything with a Trace. What I learned is that the sheer size of Trace is comically, exasperatingly large:

bigTrace = Trace[ListPlot[dat]];


ByteCount[bigTrace]

5728324392

A five and a half gigabyte trace? Really? I'll keep trying to learn more but that's just depressing. Can this be considered a bug? Someone please tell me this has been fixed after version 10.1.

Comments

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....

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