Skip to main content

plotting - Conditional options in Plot



I was hoping to incorporate an If function into a Plot option, as in


Plot[Sin[t], {t, 0, 2 Pi}, Filling -> Axis, 
FillingStyle -> If[Sin[t] > 0, LightGreen, LightRed]]

or


Plot[Sin[t], {t, 0, 2 Pi}, PlotStyle -> If[Sin[t] > 0, Dashed, Thick]]


to no avail. Is it possible to pass an If to the plot options?


I know that I can manually accomplish the effect via Piecewise, but I have more complicated applications in mind where I would not want to manually precompute the interval(s) on which my piecewise defined function would need to be defined.



Answer



Here is yet another crack at this problem.


ConditionalPlot[func_, condition_, varrange_, trueopts_, falseopts_] :=
Module[{plottrue, plotfalse},
plottrue = Plot[If[condition, func], varrange, trueopts];
plotfalse = Plot[If[Not[condition], func], varrange, falseopts];
Show[plottrue, plotfalse, PlotRange -> All]]


The first argument is the function or list of functions you want to plot. The second argument is the condition you want to apply. The third argument is the variable and range to plot in the form {x,xmin,xmax}. The third and fourth arguments are the options you apply when the condition is true or false, respectively.


For example, the plot you mentioned in your question can be had by


ConditionalPlot[Sin[x], 
Sin[x] > 0, {x, 0, 2 Pi}, {Filling -> Axis,
FillingStyle -> LightGreen, PlotStyle -> Dashed}, {Filling -> Axis,
FillingStyle -> LightRed, PlotStyle -> Thick}]

enter image description here


This is versitile, you can give it a compound condition like



ConditionalPlot[Sin[x], 
Sin[x] > .7 || Sin[x] < -.7, {x, 0, 2 Pi}, {Filling -> Axis,
FillingStyle -> LightGreen, PlotStyle -> Dashed}, {Filling -> Axis,
FillingStyle -> LightRed, PlotStyle -> Thick}]

enter image description here


You can give it multiple functions to plot


ConditionalPlot[{Sin[x], Cos[x]}, 
Sin[x] > Cos[x], {x, 0, 4 Pi},
{PlotStyle -> {Red, Thick}, Axes -> False, Frame -> True, BaseStyle -> 14},

{PlotStyle -> {Black, Thick,Dashed}}]

enter image description here


Note that any global options you want to apply to the image in general should go in the trueopts as it is given to Show first.


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}]