Skip to main content

dynamic - Graphics can't be live and editable



Is there a way to construct an interface so that the Code and the Graphics remain in sync? For example I would like to be able to drag/drop and move around different shapes in Graphic while seeing the code change. Then I would like to edit the code so the graphic changes.


Updated image to clarify interface


enter image description here


I'm aware that you can see the graphics source by hitting Ctrl+Shift+E, but this isn't ideal because the graphics source code is often in the displayform and much more verbose. Additionally it doesn't allow me to see how the code changes while I am editing the code. I'm working on some code that uses dynamic currently, but I'm not sure of the right technique just yet.


EDIT: Both the InputField and the Graphics should be live and editable. The problem currently is that the the Graphics elements can't seem to be Dynamic. If you set it as Dynamic it is impossible to re size and move the elements around.



Answer



I am still working on the code, but I have found the following solution to work the best in M8 because M8 doesn't have the Cells function. I had to use Quiet in several places to do bugs in Mathematica's internal code.


enter image description here


GetCode[] := 
  Cell[BoxData[

    StyleBox[
     DynamicBox[
      ToBoxes[Refresh[
        InputForm @@ 
         MakeExpression@
          First@First@
            Quiet[Cases[NotebookGet[EvaluationNotebook[]][[1]], 
              Cell[___, CellTags -> "MyGraphic", ___], Infinity]], 
        UpdateInterval -> 1], StandardForm]], StripOnInput -> False, 
     LineColor -> GrayLevel[0.5], FrontFaceColor -> GrayLevel[0.5], 

     BackFaceColor -> GrayLevel[0.5], GraphicsColor -> GrayLevel[0.5],
      FontColor -> GrayLevel[0.5]]], "Output", CellTags -> "MyCode"];

DynamicQ[] :=
  (Length@
     Cases[Quiet[
       Cases[NotebookGet[EvaluationNotebook[]][[1]], 
        Cell[___, CellTags -> "MyCode", ___], Infinity], 
       First::first], DynamicBox[___], Infinity] == 1);


CellPrint@
  Cell[BoxData[
    ToBoxes[Graphics[{Circle[]}, ImageSize -> {100, Automatic}]]], 
   "Output", CellTags -> "MyGraphic"];

Print[Button["Diable/Enable Dynamic", If[DynamicQ[],
      NotebookLocate["MyGraphic"];
      code = 
       InputForm @@ 
        MakeExpression@NotebookRead[EvaluationNotebook[]][[1, 1]];

      NotebookLocate["MyCode"];
      NotebookWrite[EvaluationNotebook[], 
       Cell[BoxData[ToBoxes[code]], "Input", CellTags -> "MyCode"]];
      , NotebookLocate["MyCode"];
      Quiet[NotebookWrite[EvaluationNotebook[], GetCode[]]];];

    ] Button["Update Graphics",
    NotebookLocate["MyCode"];
    code = NotebookRead[EvaluationNotebook[]][[1, 1]];
    NotebookLocate["MyGraphic"];

    NotebookWrite[EvaluationNotebook[], Cell[BoxData[ToBoxes@
        ToExpression[code]
       ], "Input", CellTags -> "MyGraphic"]
     ];]];

CellPrint@GetCode[];

Older Revisions:1st 2nd 3rd 3.1


Comments

Post a Comment

Popular posts from this blog

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...
Post a Comment
Read more