Skip to main content

Plotting piecewise function with distinct colors in each section


I have a piecewise function that I would like to plot but I was wondering if it is possible that each part of the function that is plotted when its corresponding condition is true be plotted with a different color from the other parts. That is, if I have a Piecewise function Piecewise[{{val1, cond1},{val2,cond2},{val3,cond3}}] then I want val1, val2, and val3 to be plotted with different colors so that I can differentiate each case in the plot.



Answer



Here's an alternative approach than Spartacus' answer. What he did is splitting up the piecewise function into many different functions valid in only a small domain; what I am doing here is directly plotting the piecewise function as given, while the coloring is done using ColorFunction.


I'll use the same function as Spartacus,


f = Piecewise[{{#^2, # <= 0}, {#, 0 < # <= 2}, {Log[#], 2 < #}}] &


Now let's create a ColorFunction that does the desired thing out of this. I'll do this using Part, i.e. double brackets [[ ]], which is not limited to lists only.


First, create a copy of f.



colorFunction = f;

Now we need to find out how many pieces there are in this function; for this we have to extract those into a list we can allpy Length to. Step by step:


colorFunction[[1]]


Piecewise[{{#1^2, #1 <= 0}, {#1, Inequality[0, Less, #1, LessEqual, 2]}, {Log[#1], 2 < #1}}, 0]

That's the full function body. By applying another [[1]], we can get the first argument of Piecewise:


colorFunction[[1, 1]]



{{#1^2, #1 <= 0}, {#1, 0 < #1 <= 2}, {Log[#1], 2 < #1}}

From this matrix-shaped list, we'd like to get the length, leaving us with


piecewiseParts = Length@colorFunction[[1,1]]

Alright! Now make some colors out of that. The default plot colors are stored in ColorData[1][x], where x=1,2,3,4... is the usual blue/magenta/yellowish/green and so on.


colors = ColorData[1][#] & /@ Range@piecewiseParts



{RGBColor[0.2472, 0.24, 0.6], RGBColor[0.6, 0.24, 0.442893], RGBColor[0.6, 0.547014, 0.24]}

Now we need to take these color directives and inject them into the original function (that is, the colorFunction copy I've made in the beginning), so that it replaces squares and logarithms by reds and blues. This is some more Part acrobatics:


colorFunction[[1, 1, All, 1]] = colors

Done! colorFunction is now identical to the original function f, only that the actual functions have been replaced by colors. It looks like this:


Piecewise[{{RGBColor[...], # <= 0}, {RGBColor[...], 0 < # <= 2}, {RGBColor[...], 2 < #}}] &

Now it's time to plot, see the completed code below.




f = Piecewise[{{#^2, # <= 0}, {#, 0 < # <= 2}, {Log[#], 2 < #}}] &;

colorFunction = f;
piecewiseParts = Length@colorFunction[[1, 1]];
colors = ColorData[1][#] & /@ Range@piecewiseParts;
colorFunction[[1, 1, All, 1]] = colors;

Plot[
f[x],

{x, -2, 4},
ColorFunction -> colorFunction,
ColorFunctionScaling -> False
]

final plot


(The option ColorFunctionScaling determines whether Mathematica scales the domain for the color function to $[0,1]$. Handy in some cases, not so much here, since our self-made colorFunction is constant in this domain.)


Comments

Popular posts from this blog

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...

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...

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...