Skip to main content

Using function with multiple definitions in Manipulate


I'm trying to use Manipulate to visually try out different values of lambda in a Box-Cox transformation. I've created a boxcox function with two definitions to deal with both the normal case and the case when lambda is 0:


boxcox[data_, 0] := Log[data]
boxcox[data_, l_] := (data^l - 1)/l

Then I use this function inside Manipulate but I keep getting tons of errors. It looks like Manipulate is only using the general definition and starts complaining about dividing by zero.


Manipulate[
 pdata = Partition[boxcox[data, u], 12];
 ranges = Max[#] - Min[#] & /@ pdata;

 means = Mean[#] & /@ pdata;
 mrdata = Transpose[{means, ranges}];
 mrlm = LinearModelFit[mrdata, x, x];
 Show[
  ListPlot[mrdata, Axes -> False, Frame -> True, 
   AxesOrigin -> {Automatic, 0}],
  Plot[mrlm[x], {x, Min[means], Max[means]}]
  ],
 {u, 0.00, 1.00}
 ]


Here is the data I'm using in case it matters:


data = {154., 96., 73., 49., 36., 59., 95., 169., 210., 278., 298., 245., \
200., 118., 90., 79., 78., 91., 167., 169., 289., 347., 375., 203., \
223., 104., 107., 85., 75., 99., 135., 211., 335., 460., 488., 326., \
346., 261., 224., 141., 148., 145., 223., 272., 445., 560., 612., \
467., 518., 404., 300., 210., 196., 186., 247., 343., 464., 680., \
711., 610., 613., 392., 273., 322., 189., 257., 324., 404., 677., \
858., 895., 664., 628., 308., 324., 248., 272.}

Answer




As already explained: MatchQ[0, 0.] is False.


Generally, I suggest using:


boxcox[data_, x_ /; x == 0] := Log[data]

This works even for expression that are not expressly 0 or 0., e.g.:


MatchQ[E^(I Pi/4) - (-1)^(1/4), x_ /; x == 0]


True


It also works in cases like this:


MatchQ[0.0000000000000000000, x_ /; x == 0]


True

Compare:


MatchQ[0.0000000000000000000, 0 | 0.]



False

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