4
$\begingroup$

Problem

I would like to use consecutive PlotStyles (e.g. from ColorData[...]) across different Plot/ListPlot/ListLinePlot commands. That is: each plot should pick colors from the list starting where the previous plot stopped.

In other words: I want

Show[
 Plot[{f[x], g[x]}, ... , PlotStyle -> cs],
 Plot[{h[x], i[x]}, ... , PlotStyle -> cs]
]

to be equivalent to

Plot[{f[x], g[x], h[x], i[x]}, ... , PlotStyle -> cs]

But I still want to have the same automatic color schemes for e.g. Show[Plot[...], ListPlot[...]].

First attempt

Since the PlotStyle -> ColorData[n] options seems to call ColorData[n][k] with increasing index k, I created a wrapper to make sure that k always increases:

ic[n0_Integer: 0, n1_Integer: 0, f_Function: Identity] := 
 Module[{N0 = n0, N1 = n1},
  (If[# <= N1, N0 = N0 + N1]; N1 = #; f[N0 + #]) &
 ]

Now

c = ic[];
Table[c[n], {n, 1, 5}]
Table[c[n], {n, 1, 5}]

returns

{1, 2, 3, 4, 5}
{6, 7, 8, 9, 10}

And I can do the same with indexed ColorDataFunctions like so color = ic[ColorData[n][#] &]. Repeated calls of color[1] will return consecutive colors.

But repeated Plot[..., PlotStyle -> color] will always restart from the beginning.

Improve this question
$\endgroup$
5
  • $\begingroup$ works as expected if you use PlotStyle->color[1] $\endgroup$ – kglr Jun 7 '19 at 9:46
  • $\begingroup$ @kglr Not for me. See the edit. PlotStyle -> color[1] will only give different colors for different Plot[...] calls, not for different functions within a single Plot[...]. $\endgroup$ – DLichti Jun 7 '19 at 10:05
  • $\begingroup$ With multiple functions (say 2) using PlotStyle->Table[color[1],{2}] works for both within and across plots. $\endgroup$ – kglr Jun 7 '19 at 10:16
  • $\begingroup$ @kglr Yes, it does. But it needs redundant a-priori knowledge about the number of functions in the PlotStyle option, which is one of the things I am trying to avoid. $\endgroup$ – DLichti Jun 7 '19 at 10:25
  • $\begingroup$ DLichti, i posted an alternative approach that doesn't require knowledge of how many functions/lists appear in the first argument of plot/listplot... $\endgroup$ – kglr Jun 7 '19 at 10:47
5
$\begingroup$

You can use color to define a function style and wrap the first argument of plot/chart functions with style to get the colors updated both within and across plots:

ClearAll[color, style]
color = ic[(ColorData[3][#] &)];
style = Style[#, color[1]]& /@ #&;

Show[Plot[Evaluate@style@{Sin[x], Cos[x]}, {x, 0, 2Pi}, PlotLegends -> "Expressions"], 
   Plot[Evaluate @ style @ {x Sin[x], x Cos[x]}, {x, 0, 2Pi}, PlotLegends -> "Expressions"],
   PlotRange -> All]

enter image description here

color = ic[(ColorData[3][#] &)];
Plot[Evaluate @ style @ {Sin[x], Cos[x], x Sin[x], x Cos[x]}, {x,0, 2Pi},
  PlotLegends -> "Expressions"]

enter image description here

Another example:

color = ic[(ColorData[43][#] &)];
style = Style[#,color[1]]&/@#&;

The first execution of 

Row[{Plot[Evaluate @ style @ {Sin[x], Cos[x]}, {x, 0, 2 Pi}, 
  BaseStyle -> Thick, ImageSize->200]
ListLinePlot[style @ RandomReal[1, {2, 10, 2}], 
 BaseStyle -> PointSize[Large], ImageSize->200]
PieChart[style @ Range[3], ImageSize->200]}]

gives

enter image description here

the second and third executions give

enter image description here

enter image description here

Improve this answer
$\endgroup$
2
  • $\begingroup$ Hm, I really hoped, this would be possible without messing with the first argument. Now, the style function will evaluate the first argument before Plot can extract the informations for PlotLegends -> "Expressions". Is there a way to keep the argument of style unevaluated? $\endgroup$ – DLichti Jun 7 '19 at 12:51
  • 1
    $\begingroup$ @DLichti, afaik we do need Evaluate when the function arguments are wrapped. In older versions (e.g. version 9) you could use a function for PlotStyle. So Plot[{Sin[x], Cos[x]}, {x, 0, Pi}, PlotStyle -> ({clock[1],#}&/@#&)] would give the desired result. But in versions 11+ this trick no longer works. $\endgroup$ – kglr Jun 7 '19 at 13:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.