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

Post a Comment

Popular posts from this blog

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
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