Skip to main content

performance tuning - Plotting the components of a function that returns a list in different colors without redundant evaluations of the function



I have a function f which takes a number as input, and returns a list of numbers (the length of the list is constant). f is hard to calculate (each evaluation takes a long time).


I want to plot the different components of f in different colors.


If I use this command:


Plot[f[x], {x, -2, 2}]

all the lines are drawn in the same color.
If I use this command:


Plot[{f[x][[1]], f[x][[2]], f[x][[3]]}, {x, -2, 2}]

(assuming the list has three components) the lines are drawn in different colors, but the function is called three times the necessary amount.



Note that this is a numeric function, it cannot be evaluated with a symbolic argument (i.e. the function definition begins with f[x_Real]:=), so there is no use in using Evaluate like in this question.



Answer



At risk of stating the obvious, if you are willing to give up the adaptive sampling, exclusions, etc. of Plot you could use ListLinePlot:


f[x_?NumericQ] := x + Mod[x, {1, 2, 3}]

ListLinePlot[Transpose@Table[f[x], {x, 0, 10, 0.01}], PlotStyle -> Thick]

Mathematica graphics




Better I think is to restyle the Graphics data produced by plot, as Heike did for Plotting piecewise function with distinct colors in each section and which I refactored in my answer. Applied here:



Module[{i = 1},
Plot[f[x], {x, 0, 10}, PlotStyle -> Thick] /.
x_Line :> {ColorData[1][i++], x}
]

enter image description here




Even nicer is Simon Woods' method which styles the plot while it is created, posted in answer to:



Also useful and very interesting is the solution by wxffles in the follow-up question:




Comments

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...

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...

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}]