Skip to main content

gui construction - Variable number of controls in manipulate


i want to add a variable amount of variables to manipulate, but i got stuck at one point and can't figure out a solution (even after reading here for hours...).


I have a function (e.g. here a gaussian):


f[x_, a_,μ_,σ_] := a Exp[-(x - μ)^2/(2 σ^2)]/(Sqrt[2 Pi] σ)


and define now a sum of these function, which i would like to plot using manipulate:


n=3;
test = {A[#], μ[#], σ[#], m[#]} & /@ Range[n];
func = f[x, ##] & @@@ test;

Setting initial values like this:


initialvalues = {50, 5, 20, 50, 50, 40, 100, 80, 20};
values = Partition[initialvalues, n];
rangelow = values*0.7;

rangehigh = values*1.3;

Now i want to plot this, after setting initial values with manipulate, e.g. like this:


With[{func = func},
    Manipulate[
        Show[Plot[func, {x, 0, 100}, PlotRange -> All]],
    'List of manipulate arguments'
        ]
]


With a list like this:


testlist = 
 Flatten[{{{A[#], values[[#, 1]]}, rangelow[[#, 1]], 
  rangehigh[[#, 1]]}, {{μ[#], values[[#, 2]]}, 
  rangelow[[#, 2]], 
  rangehigh[[#, 2]]}, {{σ[#], values[[#, 3]]}, 
  rangelow[[#, 3]], rangehigh[[#, 3]]}} & /@ Range[n], 1]

But I can't figure out how to remove the outer bracketsand there should be a better solution.



Answer




One can slightly rewrite your code and obtain a nice Manipulate


f[x_, a_, μ_, σ_] := a Exp[-(x - μ)^2/(2 σ^2)]/(Sqrt[2 π] σ)

n = 3;
vars = Through@{a, μ, σ}@# & /@ Range[n];
func = f[x, ##] & @@@ vars;

values = {{50, 5, 20}, {50, 50, 40}, {100, 80, 20}};
rangelow = 0.7 values;
rangehigh = 1.3 values;


With[{func = func}, 
 Manipulate[Plot[func, {x, 0, 100}, PlotRange -> All], ##] & @@ ({{#, #2}, ##3} & @@@ 
    Flatten[{vars, values, rangelow, rangehigh}, {{2, 3}}])]

enter image description here


There is two basic techniques to inject a list of variables:



  • function[..., ##] & @@ variables

  • With[{vars = Sequence @@ variables}, function[..., vars]]



Comments

Post a Comment

Popular posts from this blog

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

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

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