Skip to main content

front end - Getting the Box of a Button


I have a useful snippet of code that I use for anything in my notebook that takes a while to execute and doesn't need to be executed for subsequent cells:


buttonEvaluate[expr_] := Module[{cell}, 
   Button["Evaluate", 
     cell = Cell[BoxData[ToBoxes[ expr ]], "Output"];
     SelectionMove[ButtonNotebook[], All, ButtonCell]; 
     NotebookWrite[ButtonNotebook[], cell]

     ,
     Method -> "Queued"
   ]
];

SetAttributes[buttonEvaluate, HoldFirst];

buttonEvaluate::usage = "Returns a button, that when pressed, replaces itself with the evaluation of the passed in expression. Useful for things you don't necessarily care to recalculate when the notebook is evaluated.";

I usually use it like this:



some[expression,that,takes] + a * long / time // buttonEvaluate

...and when that cell is executed, the output is a button labelled Evaluate, which, when pressed, replaces the output cell with the result. This means I can use Evaluation->Evaluate Notebook, and it effectively skips over some cells until I explicitly choose to evaluate them.


However, I've run into an issue when I try to use it inside an expression, because the perfectionist in me wants to be able to do something like:


Row[{slowThing[] // buttonEvaluate, fastThing[]}]

However, the Button's action is to replace the ButtonCell with the results, which inadvertently replaces everything, not just the Button's Box. So, two questions:



  1. Is there any way to select the Box of the Button?

  2. Will replacing that selection using NotebookWrite work how I want it to? I can imagine it might play havoc with Mathematica's formatting.




Answer



V10 edit


As of V10 it is easier to do that with a help of EvaluationBox[]. Your second concern is solved by temporarily changing DefaultDuplicateCellStyle. 78417


ButtonHold ~ SetAttributes ~ HoldFirst;

ButtonHold[expr_] := Button[
   Tooltip["Evaluate", HoldForm[expr]]
   ,
   Module[{nb = EvaluationNotebook[], pre, opt = DefaultDuplicateCellStyle},

     pre = Options[nb, opt];
     SetOptions[nb, opt -> "Output"];
     NotebookWrite[EvaluationBox[], ToBoxes @ expr];
     SetOptions[nb, pre]
   ]
   ,
   Method -> "Queued"
];

Grid[{{1, Button[1]}, {ButtonHold[Plot[Sin[x], {x, 0, 1}]], 

  4}, {ButtonHold[Plot[Cos[x], {x, 0, 1}]], 10}}]

enter image description here



Old solution


Not pretty but works. The trick is to inject Unique string to the box so it can be found later in whatever you wish to have in that cell.


Cell expression with replaced BouttonBox is then overwritting button's parent cell.


buttonEvaluate2 = Function[expr,

   Button["Evaluate", 

          #;
          SelectionMove[EvaluationCell[], All, Cell]; 
          NotebookWrite[ EvaluationNotebook[],
                         NotebookRead[EvaluationCell[]
                              ] /. b_ButtonBox /; MemberQ[b, #, {3}] :> ToBoxes@expr
                         ];
          , Method -> "Queued"
         ] &[ ToString @ Unique["x"] ]
   ,
   HoldFirst];



test


Grid[{
  {1, Button[1]},
  {buttonEvaluate2[Plot[Sin[x], {x, 0, 1}]], 4},
  {buttonEvaluate2[Plot[Cos[x], {x, 0, 1}]], 10}
  }]

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