Skip to main content

wolfram cloud - CloudDeploy a Manipulate with dependent functions


I don't seem to understand CloudDeploy; it seems like it deploys a function, but not its dependent expressions. How should I deploy an expression that may have dependencies previously in the notebook?


For example, I evaluated the following expression:


f=Sin[x];CloudDeploy[Manipulate[Plot[f,{x,0,t}],{t,1,15}]]


I then visited the deployed page, and got a plot with no function.


This is a simplified version; in more complex ones, such as when PlotRange depends on something previously defined, I can see error messages.


It seems that CloudDeploy's deployed object definition isn't closed over the set of definitions in effect. However, the docs (at ref/CloudDeploy.html) specify that:



CloudDeploy[expr,…] automatically deploys all definitions needed to evaluate expr, much like CloudSave.



Am I misunderstanding something here?



Answer



The quick fix is to use Manipulate's SaveDefinitions, making it responsible for storing dependencies:


f = Sin[x];

CloudDeploy[
Manipulate[Plot[f, {x, 0, t}], {t, 1, 15}, SaveDefinitions -> True]
]

I don't know if that is expected, maybe Documentation sticks to the wording precisely here. So it happens because f is not needed to evaluate Manipulate (it is needed to display it):


InputForm @ Manipulate[Plot[f, {x, 0, t}], {t, 1, 15}]

in contrary to e.g. InputForm @ Grid[f].


On the other hand it does not make much sense since CloudDeploy does not Hold its arguments which means expr is evaluated before sending, which means no additional definition will ever be needed if we stick to the wording...


So maybe it is worth reporting?





Compare also:


CloudDeploy[Delayed[f]]

CloudDeploy[Dynamic[f]]

Comments

Popular posts from this blog

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 - 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 - 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.