Skip to main content

Is there any way to allow a message to appear more than three times before General::stop shuts it off?


I have some Mathematica code that analyzes data in a database and uses Message to report when expected data is missing. By default, Mathematica will allow this message to appear only three times before it gets shut off, but I'd like this message to appear as often as the errors do.



The solutions I can think of are


Off[General::stop];

but this kills limits for all error messages, not just this one,


or just to switch to Print, which seems inelegant.


Is there some way to allow this one message to appear more than three times, while leaving behavior unchanged for the rest?



Answer



You can do something like this:


resetMessages[symbol_] := 
 With[{mysymbol = symbol}, 

  Unprotect[$MessageList]; $MessageList = 
   DeleteCases[$MessageList, HoldForm[MessageName[mysymbol, _]]]; 
  Protect[$MessageList];]

And you will have to call it after each function...


Sqrt[a, b, c, d]; Exp[a, b]; resetMessages[Exp]; Sqrt[a, b, c, d]; \
Exp[a, b]; resetMessages[Exp]; Sqrt[a, b, c, d]; Exp[a, b]; \
resetMessages[Exp]; Sqrt[a, b, c, d]; Exp[a, b]; resetMessages[Exp]; \
Sqrt[a, b, c, d]; Exp[a, b]; resetMessages[Exp]; Sqrt[a, b, c, d]; \
Exp[a, b]; resetMessages[Exp]; Sqrt[a, b, c, d]; Exp[a, b]; \

resetMessages[Exp];

Sqrt::argx: Sqrt called with 4 arguments; 1 argument is expected. >>

Exp::argx: Exp called with 2 arguments; 1 argument is expected. >>

Sqrt::argx: Sqrt called with 4 arguments; 1 argument is expected. >>

Exp::argx: Exp called with 2 arguments; 1 argument is expected. >>


Sqrt::argx: Sqrt called with 4 arguments; 1 argument is expected. >>

General::stop: Further output of Sqrt::argx will be suppressed during this calculation. >>

Exp::argx: Exp called with 2 arguments; 1 argument is expected. >>

Exp::argx: Exp called with 2 arguments; 1 argument is expected. >>

Exp::argx: Exp called with 2 arguments; 1 argument is expected. >>


Exp::argx: Exp called with 2 arguments; 1 argument is expected. >>

Exp::argx: Exp called with 2 arguments; 1 argument is expected. >>

Comments

Post a Comment

Popular posts from this blog

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}]
Post a Comment
Read more

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}]
Post a Comment
Read more

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....
Post a Comment
Read more