Recolour existing plotsConsecutive `PlotStyle` across multiple plotsPreparing 2d plots for...

What is the difference between a historical drama and a period drama?

How frequently do Russian people still refer to others by their patronymic (отчество)?

Use real text instead of lipsum in moderncv quote alignment

Should I cross-validate metrics that were not optimised?

What can a novel do that film and TV cannot?

Was Wolfgang Unzicker the last Amateur GM?

My players like to search everything. What do they find?

Do the 26 richest billionaires own as much wealth as the poorest 3.8 billion people?

What is meaning of 4 letter acronyms in Roman names like Titus Flavius T. f. T. n. Sabinus?

gzip compress a local folder and extract it to remote server

Finding integer database columns that may have their data type changed to reduce size

Are there advantages in writing by hand over typing out a story?

Show that there are infinitely more problems than we will ever be able to compute

Can you move between the attacks of a Twinned Booming Blade

Is my background sufficient to start Quantum Computing

Does a reference have a storage location?

When you're given a degree sequence, what is the method to draw a graph which has that degree sequence?

My mother co-signed for my car. Can she take it away from me if I am the one making car payments?

List of Implementations for common OR problems

Phrasing "it says" or "it reads"

Does this circuit have marginal voltage level problem?

How can I know (without going to the station) if RATP is offering the Anti Pollution tickets?

What do you call the motor that fuels the movement of a robotic arm?

What is -(-2,3,4)?



Recolour existing plots


Consecutive `PlotStyle` across multiple plotsPreparing 2d plots for publicationControlling what is plotted in a multi-curve plotHow to change the number of markers in Plots?Plotting: DensityPlot produces artifactsList of Inequalities in RegionPlot with different colorsUsing different style on segments of a plotHow to plot a pair of functions as solid and dashed, but a list of those pairs as different colors, in one plot?Why could not FindAllCrossings2D find the other intersection of contour plots?Plot Function is not Coloring different plots correctlyListPlot in a Table with different colours and labels






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}







3












$begingroup$


Context



More often than not I end up doing plots one after the other so that I have a set of plots with the same colour style.



When I want to make it more publishable ready, I would like to reassign colours to each line.




How to assign colours from a given style to existing sets of plots?




Example



pl1= Plot[Sin[x],{x,0,4Pi}];
pl2= Plot[Cos[2x],{x,0,4Pi}];
Show[pl1,pl2];


Mathematica graphics



Attempt



I while back I wrote the following function



Clear[ShowColor];
ShowColor[list___]:=ShowColor[{list}]/; Length[{list}]>1;
ShowColor[list_,ColorRange->color__,opt___]:= Module[{len=Length[list]},
Table[list[[i]] /. RGBColor[_,_,_]->
GradientColor[color][(i-1)/(len-1)],{i,len}]//Show[#,opt]&]
ShowColor[list_,opt___]:= Module[{len=Length[list]},
Table[list[[i]] /. RGBColor[_,_,_]->
GradientColor[ColorData[10] /@ Range[10]][(i-1)/(len-1)],{i,len}]//Show[#,opt]&]


which uses the GradientColor Package,
so that



ShowColor[{pl1,pl2}]


produces



Mathematica graphics



But I am left with the impression that it could be done more elegantly and generally with the modern version of Mathematica, making use of the set of default styles
and working in harmony with other features.



Also, my implementation is not very robust. For instance,



 Show[pl1, pl2] // ShowColor 


fails.



What would be great would be to have a function which
e.g. would take standard Options such as



 ShowColor[plots,PlotStyle-> ColorData[10]]


or



  ShowColor[plots,PlotStyle-> Directive[{Dashed,Blue}]]



Any suggestion on how to make this as generic as possible?




Thanks!










share|improve this question











$endgroup$












  • $begingroup$
    you might find this interesting.
    $endgroup$
    – kglr
    7 hours ago












  • $begingroup$
    thanks for the link
    $endgroup$
    – chris
    7 hours ago


















3












$begingroup$


Context



More often than not I end up doing plots one after the other so that I have a set of plots with the same colour style.



When I want to make it more publishable ready, I would like to reassign colours to each line.




How to assign colours from a given style to existing sets of plots?




Example



pl1= Plot[Sin[x],{x,0,4Pi}];
pl2= Plot[Cos[2x],{x,0,4Pi}];
Show[pl1,pl2];


Mathematica graphics



Attempt



I while back I wrote the following function



Clear[ShowColor];
ShowColor[list___]:=ShowColor[{list}]/; Length[{list}]>1;
ShowColor[list_,ColorRange->color__,opt___]:= Module[{len=Length[list]},
Table[list[[i]] /. RGBColor[_,_,_]->
GradientColor[color][(i-1)/(len-1)],{i,len}]//Show[#,opt]&]
ShowColor[list_,opt___]:= Module[{len=Length[list]},
Table[list[[i]] /. RGBColor[_,_,_]->
GradientColor[ColorData[10] /@ Range[10]][(i-1)/(len-1)],{i,len}]//Show[#,opt]&]


which uses the GradientColor Package,
so that



ShowColor[{pl1,pl2}]


produces



Mathematica graphics



But I am left with the impression that it could be done more elegantly and generally with the modern version of Mathematica, making use of the set of default styles
and working in harmony with other features.



Also, my implementation is not very robust. For instance,



 Show[pl1, pl2] // ShowColor 


fails.



What would be great would be to have a function which
e.g. would take standard Options such as



 ShowColor[plots,PlotStyle-> ColorData[10]]


or



  ShowColor[plots,PlotStyle-> Directive[{Dashed,Blue}]]



Any suggestion on how to make this as generic as possible?




Thanks!










share|improve this question











$endgroup$












  • $begingroup$
    you might find this interesting.
    $endgroup$
    – kglr
    7 hours ago












  • $begingroup$
    thanks for the link
    $endgroup$
    – chris
    7 hours ago














3












3








3


1



$begingroup$


Context



More often than not I end up doing plots one after the other so that I have a set of plots with the same colour style.



When I want to make it more publishable ready, I would like to reassign colours to each line.




How to assign colours from a given style to existing sets of plots?




Example



pl1= Plot[Sin[x],{x,0,4Pi}];
pl2= Plot[Cos[2x],{x,0,4Pi}];
Show[pl1,pl2];


Mathematica graphics



Attempt



I while back I wrote the following function



Clear[ShowColor];
ShowColor[list___]:=ShowColor[{list}]/; Length[{list}]>1;
ShowColor[list_,ColorRange->color__,opt___]:= Module[{len=Length[list]},
Table[list[[i]] /. RGBColor[_,_,_]->
GradientColor[color][(i-1)/(len-1)],{i,len}]//Show[#,opt]&]
ShowColor[list_,opt___]:= Module[{len=Length[list]},
Table[list[[i]] /. RGBColor[_,_,_]->
GradientColor[ColorData[10] /@ Range[10]][(i-1)/(len-1)],{i,len}]//Show[#,opt]&]


which uses the GradientColor Package,
so that



ShowColor[{pl1,pl2}]


produces



Mathematica graphics



But I am left with the impression that it could be done more elegantly and generally with the modern version of Mathematica, making use of the set of default styles
and working in harmony with other features.



Also, my implementation is not very robust. For instance,



 Show[pl1, pl2] // ShowColor 


fails.



What would be great would be to have a function which
e.g. would take standard Options such as



 ShowColor[plots,PlotStyle-> ColorData[10]]


or



  ShowColor[plots,PlotStyle-> Directive[{Dashed,Blue}]]



Any suggestion on how to make this as generic as possible?




Thanks!










share|improve this question











$endgroup$




Context



More often than not I end up doing plots one after the other so that I have a set of plots with the same colour style.



When I want to make it more publishable ready, I would like to reassign colours to each line.




How to assign colours from a given style to existing sets of plots?




Example



pl1= Plot[Sin[x],{x,0,4Pi}];
pl2= Plot[Cos[2x],{x,0,4Pi}];
Show[pl1,pl2];


Mathematica graphics



Attempt



I while back I wrote the following function



Clear[ShowColor];
ShowColor[list___]:=ShowColor[{list}]/; Length[{list}]>1;
ShowColor[list_,ColorRange->color__,opt___]:= Module[{len=Length[list]},
Table[list[[i]] /. RGBColor[_,_,_]->
GradientColor[color][(i-1)/(len-1)],{i,len}]//Show[#,opt]&]
ShowColor[list_,opt___]:= Module[{len=Length[list]},
Table[list[[i]] /. RGBColor[_,_,_]->
GradientColor[ColorData[10] /@ Range[10]][(i-1)/(len-1)],{i,len}]//Show[#,opt]&]


which uses the GradientColor Package,
so that



ShowColor[{pl1,pl2}]


produces



Mathematica graphics



But I am left with the impression that it could be done more elegantly and generally with the modern version of Mathematica, making use of the set of default styles
and working in harmony with other features.



Also, my implementation is not very robust. For instance,



 Show[pl1, pl2] // ShowColor 


fails.



What would be great would be to have a function which
e.g. would take standard Options such as



 ShowColor[plots,PlotStyle-> ColorData[10]]


or



  ShowColor[plots,PlotStyle-> Directive[{Dashed,Blue}]]



Any suggestion on how to make this as generic as possible?




Thanks!







plotting graphics






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 8 hours ago







chris

















asked 9 hours ago









chrischris

12.5k4 gold badges45 silver badges118 bronze badges




12.5k4 gold badges45 silver badges118 bronze badges












  • $begingroup$
    you might find this interesting.
    $endgroup$
    – kglr
    7 hours ago












  • $begingroup$
    thanks for the link
    $endgroup$
    – chris
    7 hours ago


















  • $begingroup$
    you might find this interesting.
    $endgroup$
    – kglr
    7 hours ago












  • $begingroup$
    thanks for the link
    $endgroup$
    – chris
    7 hours ago
















$begingroup$
you might find this interesting.
$endgroup$
– kglr
7 hours ago






$begingroup$
you might find this interesting.
$endgroup$
– kglr
7 hours ago














$begingroup$
thanks for the link
$endgroup$
– chris
7 hours ago




$begingroup$
thanks for the link
$endgroup$
– chris
7 hours ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

Using DLichti's ingenious idea / function from this q/a:



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


to define a function color which increments the color every time it is invoked as color[1]:



ClearAll[color, reColor]
color = dLichtiIncrement[(ColorData[97][#] &)];
reColor[] = # /. _?ColorQ :> color[1] &;
reColor[_] := Module[{}, ClearAll[color];
color = dLichtiIncrement[(ColorData[97][#] &)]; reColor[]]


Examples:



pl1 = Plot[Sin[x], {x, 0, 4 Pi}];
pl2 = Plot[Cos[2 x], {x, 0, 4 Pi}];
Show[pl1, pl2]//reColor[]


enter image description here



ContourPlot[Cos[x] + Cos[y], {x, 0, 4Pi}, {y, 0, 4Pi}] // reColor[]


enter image description here



Plot[{x Sin[x], x Cos[x], Sin[x Cos[x]]}, {x, 0, 2 Pi}, 
PlotTheme -> "Monochrome", Filling -> Axis, FillingStyle -> Opacity[.5]] // reColor[]


enter image description here



ContourPlot[Evaluate[Sum[Sin[RandomReal[5, 2].{x, y}], {5}]], {x, 0, 5}, {y, 0, 5},
PlotTheme -> "Monochrome"] // reColor[]


enter image description here



 ContourPlot[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi},
ContourShading -> False] // reColor[]


enter image description here



You can also use color[1] in setting ChartStyle/PlotStyle:



BarChart[{{1, 2, 3}, {1, 3, 2}}, ChartStyle -> Table[color[1], 3]]


enter image description here



Using reColor[blah] @ Red resets color[1] to its initial state:



 reColor[blah] @ Red == ColorData[97][1]



True







share|improve this answer











$endgroup$













  • $begingroup$
    This is almost perfect: would it be possible to give reColor an argument to allow for the colours to reincrement from the beginning? I.e. plots//reColor[] would yield the same sets of colours ?
    $endgroup$
    – chris
    6 hours ago










  • $begingroup$
    @chris, i made small changes to allow resetting: regular call is reColor[] and reColor[anything] resets color to its initial definition.
    $endgroup$
    – kglr
    5 hours ago






  • 1




    $begingroup$
    Great ! I modified it slightly to choose the ColourTable: ClearAll[color, reColor] color = dLichtiIncrement[(ColorData[10][#] &)]; reColor[] = # /. _?ColorQ :> color[1] &; reColor[val_: 10] := Module[{}, ClearAll[color]; color = dLichtiIncrement[(ColorData[val][#] &)]; reColor[]]
    $endgroup$
    – chris
    4 hours ago



















2












$begingroup$

A simple way is to pass the coordinates in the plots to ListLinePlot.



recolor[plot_, opts___] := ListLinePlot[
Cases[plot, Line[coords_] :> coords, Infinity],
opts
]

Show[
recolor[pl1, PlotStyle -> Directive[Blue, Dashed]],
recolor[pl2, PlotStyle -> Red]
]


Mathematica graphics



It can also be used to recolor already combined plots:



recolor[
Plot[{Sin[x], Cos[2 x]}, {x, 0, 4 Pi}],
PlotStyle -> {
Directive[Blue, Dashed],
Red
}]


Mathematica graphics



And it also works on this:



recolor[Show[pl1, pl2],
PlotStyle -> {
Directive[Blue, Dashed],
Red
}]





share|improve this answer











$endgroup$













  • $begingroup$
    nice! I will keep the options opened in case someone comes up with an even more generic answer which would work with e.g. contour plots as well?
    $endgroup$
    – chris
    8 hours ago










  • $begingroup$
    @chris that's always a good idea, you never know what someone might come up with.
    $endgroup$
    – C. E.
    8 hours ago










  • $begingroup$
    So the next challenge would be to make it work with plots = {ContourPlot[Sin[ x y], {x, -1, 1}, {y, -1, 1}, ContourShading -> False], ContourPlot[Cos[ x - y], {x, -1, 1}, {y, -1, 1}, ContourShading -> False]}; but feel free to ignore this 'new' question.
    $endgroup$
    – chris
    7 hours ago
















Your Answer








StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f201318%2frecolour-existing-plots%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Using DLichti's ingenious idea / function from this q/a:



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


to define a function color which increments the color every time it is invoked as color[1]:



ClearAll[color, reColor]
color = dLichtiIncrement[(ColorData[97][#] &)];
reColor[] = # /. _?ColorQ :> color[1] &;
reColor[_] := Module[{}, ClearAll[color];
color = dLichtiIncrement[(ColorData[97][#] &)]; reColor[]]


Examples:



pl1 = Plot[Sin[x], {x, 0, 4 Pi}];
pl2 = Plot[Cos[2 x], {x, 0, 4 Pi}];
Show[pl1, pl2]//reColor[]


enter image description here



ContourPlot[Cos[x] + Cos[y], {x, 0, 4Pi}, {y, 0, 4Pi}] // reColor[]


enter image description here



Plot[{x Sin[x], x Cos[x], Sin[x Cos[x]]}, {x, 0, 2 Pi}, 
PlotTheme -> "Monochrome", Filling -> Axis, FillingStyle -> Opacity[.5]] // reColor[]


enter image description here



ContourPlot[Evaluate[Sum[Sin[RandomReal[5, 2].{x, y}], {5}]], {x, 0, 5}, {y, 0, 5},
PlotTheme -> "Monochrome"] // reColor[]


enter image description here



 ContourPlot[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi},
ContourShading -> False] // reColor[]


enter image description here



You can also use color[1] in setting ChartStyle/PlotStyle:



BarChart[{{1, 2, 3}, {1, 3, 2}}, ChartStyle -> Table[color[1], 3]]


enter image description here



Using reColor[blah] @ Red resets color[1] to its initial state:



 reColor[blah] @ Red == ColorData[97][1]



True







share|improve this answer











$endgroup$













  • $begingroup$
    This is almost perfect: would it be possible to give reColor an argument to allow for the colours to reincrement from the beginning? I.e. plots//reColor[] would yield the same sets of colours ?
    $endgroup$
    – chris
    6 hours ago










  • $begingroup$
    @chris, i made small changes to allow resetting: regular call is reColor[] and reColor[anything] resets color to its initial definition.
    $endgroup$
    – kglr
    5 hours ago






  • 1




    $begingroup$
    Great ! I modified it slightly to choose the ColourTable: ClearAll[color, reColor] color = dLichtiIncrement[(ColorData[10][#] &)]; reColor[] = # /. _?ColorQ :> color[1] &; reColor[val_: 10] := Module[{}, ClearAll[color]; color = dLichtiIncrement[(ColorData[val][#] &)]; reColor[]]
    $endgroup$
    – chris
    4 hours ago
















3












$begingroup$

Using DLichti's ingenious idea / function from this q/a:



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


to define a function color which increments the color every time it is invoked as color[1]:



ClearAll[color, reColor]
color = dLichtiIncrement[(ColorData[97][#] &)];
reColor[] = # /. _?ColorQ :> color[1] &;
reColor[_] := Module[{}, ClearAll[color];
color = dLichtiIncrement[(ColorData[97][#] &)]; reColor[]]


Examples:



pl1 = Plot[Sin[x], {x, 0, 4 Pi}];
pl2 = Plot[Cos[2 x], {x, 0, 4 Pi}];
Show[pl1, pl2]//reColor[]


enter image description here



ContourPlot[Cos[x] + Cos[y], {x, 0, 4Pi}, {y, 0, 4Pi}] // reColor[]


enter image description here



Plot[{x Sin[x], x Cos[x], Sin[x Cos[x]]}, {x, 0, 2 Pi}, 
PlotTheme -> "Monochrome", Filling -> Axis, FillingStyle -> Opacity[.5]] // reColor[]


enter image description here



ContourPlot[Evaluate[Sum[Sin[RandomReal[5, 2].{x, y}], {5}]], {x, 0, 5}, {y, 0, 5},
PlotTheme -> "Monochrome"] // reColor[]


enter image description here



 ContourPlot[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi},
ContourShading -> False] // reColor[]


enter image description here



You can also use color[1] in setting ChartStyle/PlotStyle:



BarChart[{{1, 2, 3}, {1, 3, 2}}, ChartStyle -> Table[color[1], 3]]


enter image description here



Using reColor[blah] @ Red resets color[1] to its initial state:



 reColor[blah] @ Red == ColorData[97][1]



True







share|improve this answer











$endgroup$













  • $begingroup$
    This is almost perfect: would it be possible to give reColor an argument to allow for the colours to reincrement from the beginning? I.e. plots//reColor[] would yield the same sets of colours ?
    $endgroup$
    – chris
    6 hours ago










  • $begingroup$
    @chris, i made small changes to allow resetting: regular call is reColor[] and reColor[anything] resets color to its initial definition.
    $endgroup$
    – kglr
    5 hours ago






  • 1




    $begingroup$
    Great ! I modified it slightly to choose the ColourTable: ClearAll[color, reColor] color = dLichtiIncrement[(ColorData[10][#] &)]; reColor[] = # /. _?ColorQ :> color[1] &; reColor[val_: 10] := Module[{}, ClearAll[color]; color = dLichtiIncrement[(ColorData[val][#] &)]; reColor[]]
    $endgroup$
    – chris
    4 hours ago














3












3








3





$begingroup$

Using DLichti's ingenious idea / function from this q/a:



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


to define a function color which increments the color every time it is invoked as color[1]:



ClearAll[color, reColor]
color = dLichtiIncrement[(ColorData[97][#] &)];
reColor[] = # /. _?ColorQ :> color[1] &;
reColor[_] := Module[{}, ClearAll[color];
color = dLichtiIncrement[(ColorData[97][#] &)]; reColor[]]


Examples:



pl1 = Plot[Sin[x], {x, 0, 4 Pi}];
pl2 = Plot[Cos[2 x], {x, 0, 4 Pi}];
Show[pl1, pl2]//reColor[]


enter image description here



ContourPlot[Cos[x] + Cos[y], {x, 0, 4Pi}, {y, 0, 4Pi}] // reColor[]


enter image description here



Plot[{x Sin[x], x Cos[x], Sin[x Cos[x]]}, {x, 0, 2 Pi}, 
PlotTheme -> "Monochrome", Filling -> Axis, FillingStyle -> Opacity[.5]] // reColor[]


enter image description here



ContourPlot[Evaluate[Sum[Sin[RandomReal[5, 2].{x, y}], {5}]], {x, 0, 5}, {y, 0, 5},
PlotTheme -> "Monochrome"] // reColor[]


enter image description here



 ContourPlot[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi},
ContourShading -> False] // reColor[]


enter image description here



You can also use color[1] in setting ChartStyle/PlotStyle:



BarChart[{{1, 2, 3}, {1, 3, 2}}, ChartStyle -> Table[color[1], 3]]


enter image description here



Using reColor[blah] @ Red resets color[1] to its initial state:



 reColor[blah] @ Red == ColorData[97][1]



True







share|improve this answer











$endgroup$



Using DLichti's ingenious idea / function from this q/a:



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


to define a function color which increments the color every time it is invoked as color[1]:



ClearAll[color, reColor]
color = dLichtiIncrement[(ColorData[97][#] &)];
reColor[] = # /. _?ColorQ :> color[1] &;
reColor[_] := Module[{}, ClearAll[color];
color = dLichtiIncrement[(ColorData[97][#] &)]; reColor[]]


Examples:



pl1 = Plot[Sin[x], {x, 0, 4 Pi}];
pl2 = Plot[Cos[2 x], {x, 0, 4 Pi}];
Show[pl1, pl2]//reColor[]


enter image description here



ContourPlot[Cos[x] + Cos[y], {x, 0, 4Pi}, {y, 0, 4Pi}] // reColor[]


enter image description here



Plot[{x Sin[x], x Cos[x], Sin[x Cos[x]]}, {x, 0, 2 Pi}, 
PlotTheme -> "Monochrome", Filling -> Axis, FillingStyle -> Opacity[.5]] // reColor[]


enter image description here



ContourPlot[Evaluate[Sum[Sin[RandomReal[5, 2].{x, y}], {5}]], {x, 0, 5}, {y, 0, 5},
PlotTheme -> "Monochrome"] // reColor[]


enter image description here



 ContourPlot[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi},
ContourShading -> False] // reColor[]


enter image description here



You can also use color[1] in setting ChartStyle/PlotStyle:



BarChart[{{1, 2, 3}, {1, 3, 2}}, ChartStyle -> Table[color[1], 3]]


enter image description here



Using reColor[blah] @ Red resets color[1] to its initial state:



 reColor[blah] @ Red == ColorData[97][1]



True








share|improve this answer














share|improve this answer



share|improve this answer








edited 5 hours ago

























answered 6 hours ago









kglrkglr

202k10 gold badges230 silver badges461 bronze badges




202k10 gold badges230 silver badges461 bronze badges












  • $begingroup$
    This is almost perfect: would it be possible to give reColor an argument to allow for the colours to reincrement from the beginning? I.e. plots//reColor[] would yield the same sets of colours ?
    $endgroup$
    – chris
    6 hours ago










  • $begingroup$
    @chris, i made small changes to allow resetting: regular call is reColor[] and reColor[anything] resets color to its initial definition.
    $endgroup$
    – kglr
    5 hours ago






  • 1




    $begingroup$
    Great ! I modified it slightly to choose the ColourTable: ClearAll[color, reColor] color = dLichtiIncrement[(ColorData[10][#] &)]; reColor[] = # /. _?ColorQ :> color[1] &; reColor[val_: 10] := Module[{}, ClearAll[color]; color = dLichtiIncrement[(ColorData[val][#] &)]; reColor[]]
    $endgroup$
    – chris
    4 hours ago


















  • $begingroup$
    This is almost perfect: would it be possible to give reColor an argument to allow for the colours to reincrement from the beginning? I.e. plots//reColor[] would yield the same sets of colours ?
    $endgroup$
    – chris
    6 hours ago










  • $begingroup$
    @chris, i made small changes to allow resetting: regular call is reColor[] and reColor[anything] resets color to its initial definition.
    $endgroup$
    – kglr
    5 hours ago






  • 1




    $begingroup$
    Great ! I modified it slightly to choose the ColourTable: ClearAll[color, reColor] color = dLichtiIncrement[(ColorData[10][#] &)]; reColor[] = # /. _?ColorQ :> color[1] &; reColor[val_: 10] := Module[{}, ClearAll[color]; color = dLichtiIncrement[(ColorData[val][#] &)]; reColor[]]
    $endgroup$
    – chris
    4 hours ago
















$begingroup$
This is almost perfect: would it be possible to give reColor an argument to allow for the colours to reincrement from the beginning? I.e. plots//reColor[] would yield the same sets of colours ?
$endgroup$
– chris
6 hours ago




$begingroup$
This is almost perfect: would it be possible to give reColor an argument to allow for the colours to reincrement from the beginning? I.e. plots//reColor[] would yield the same sets of colours ?
$endgroup$
– chris
6 hours ago












$begingroup$
@chris, i made small changes to allow resetting: regular call is reColor[] and reColor[anything] resets color to its initial definition.
$endgroup$
– kglr
5 hours ago




$begingroup$
@chris, i made small changes to allow resetting: regular call is reColor[] and reColor[anything] resets color to its initial definition.
$endgroup$
– kglr
5 hours ago




1




1




$begingroup$
Great ! I modified it slightly to choose the ColourTable: ClearAll[color, reColor] color = dLichtiIncrement[(ColorData[10][#] &)]; reColor[] = # /. _?ColorQ :> color[1] &; reColor[val_: 10] := Module[{}, ClearAll[color]; color = dLichtiIncrement[(ColorData[val][#] &)]; reColor[]]
$endgroup$
– chris
4 hours ago




$begingroup$
Great ! I modified it slightly to choose the ColourTable: ClearAll[color, reColor] color = dLichtiIncrement[(ColorData[10][#] &)]; reColor[] = # /. _?ColorQ :> color[1] &; reColor[val_: 10] := Module[{}, ClearAll[color]; color = dLichtiIncrement[(ColorData[val][#] &)]; reColor[]]
$endgroup$
– chris
4 hours ago













2












$begingroup$

A simple way is to pass the coordinates in the plots to ListLinePlot.



recolor[plot_, opts___] := ListLinePlot[
Cases[plot, Line[coords_] :> coords, Infinity],
opts
]

Show[
recolor[pl1, PlotStyle -> Directive[Blue, Dashed]],
recolor[pl2, PlotStyle -> Red]
]


Mathematica graphics



It can also be used to recolor already combined plots:



recolor[
Plot[{Sin[x], Cos[2 x]}, {x, 0, 4 Pi}],
PlotStyle -> {
Directive[Blue, Dashed],
Red
}]


Mathematica graphics



And it also works on this:



recolor[Show[pl1, pl2],
PlotStyle -> {
Directive[Blue, Dashed],
Red
}]





share|improve this answer











$endgroup$













  • $begingroup$
    nice! I will keep the options opened in case someone comes up with an even more generic answer which would work with e.g. contour plots as well?
    $endgroup$
    – chris
    8 hours ago










  • $begingroup$
    @chris that's always a good idea, you never know what someone might come up with.
    $endgroup$
    – C. E.
    8 hours ago










  • $begingroup$
    So the next challenge would be to make it work with plots = {ContourPlot[Sin[ x y], {x, -1, 1}, {y, -1, 1}, ContourShading -> False], ContourPlot[Cos[ x - y], {x, -1, 1}, {y, -1, 1}, ContourShading -> False]}; but feel free to ignore this 'new' question.
    $endgroup$
    – chris
    7 hours ago


















2












$begingroup$

A simple way is to pass the coordinates in the plots to ListLinePlot.



recolor[plot_, opts___] := ListLinePlot[
Cases[plot, Line[coords_] :> coords, Infinity],
opts
]

Show[
recolor[pl1, PlotStyle -> Directive[Blue, Dashed]],
recolor[pl2, PlotStyle -> Red]
]


Mathematica graphics



It can also be used to recolor already combined plots:



recolor[
Plot[{Sin[x], Cos[2 x]}, {x, 0, 4 Pi}],
PlotStyle -> {
Directive[Blue, Dashed],
Red
}]


Mathematica graphics



And it also works on this:



recolor[Show[pl1, pl2],
PlotStyle -> {
Directive[Blue, Dashed],
Red
}]





share|improve this answer











$endgroup$













  • $begingroup$
    nice! I will keep the options opened in case someone comes up with an even more generic answer which would work with e.g. contour plots as well?
    $endgroup$
    – chris
    8 hours ago










  • $begingroup$
    @chris that's always a good idea, you never know what someone might come up with.
    $endgroup$
    – C. E.
    8 hours ago










  • $begingroup$
    So the next challenge would be to make it work with plots = {ContourPlot[Sin[ x y], {x, -1, 1}, {y, -1, 1}, ContourShading -> False], ContourPlot[Cos[ x - y], {x, -1, 1}, {y, -1, 1}, ContourShading -> False]}; but feel free to ignore this 'new' question.
    $endgroup$
    – chris
    7 hours ago
















2












2








2





$begingroup$

A simple way is to pass the coordinates in the plots to ListLinePlot.



recolor[plot_, opts___] := ListLinePlot[
Cases[plot, Line[coords_] :> coords, Infinity],
opts
]

Show[
recolor[pl1, PlotStyle -> Directive[Blue, Dashed]],
recolor[pl2, PlotStyle -> Red]
]


Mathematica graphics



It can also be used to recolor already combined plots:



recolor[
Plot[{Sin[x], Cos[2 x]}, {x, 0, 4 Pi}],
PlotStyle -> {
Directive[Blue, Dashed],
Red
}]


Mathematica graphics



And it also works on this:



recolor[Show[pl1, pl2],
PlotStyle -> {
Directive[Blue, Dashed],
Red
}]





share|improve this answer











$endgroup$



A simple way is to pass the coordinates in the plots to ListLinePlot.



recolor[plot_, opts___] := ListLinePlot[
Cases[plot, Line[coords_] :> coords, Infinity],
opts
]

Show[
recolor[pl1, PlotStyle -> Directive[Blue, Dashed]],
recolor[pl2, PlotStyle -> Red]
]


Mathematica graphics



It can also be used to recolor already combined plots:



recolor[
Plot[{Sin[x], Cos[2 x]}, {x, 0, 4 Pi}],
PlotStyle -> {
Directive[Blue, Dashed],
Red
}]


Mathematica graphics



And it also works on this:



recolor[Show[pl1, pl2],
PlotStyle -> {
Directive[Blue, Dashed],
Red
}]






share|improve this answer














share|improve this answer



share|improve this answer








edited 8 hours ago

























answered 8 hours ago









C. E.C. E.

52.6k3 gold badges102 silver badges210 bronze badges




52.6k3 gold badges102 silver badges210 bronze badges












  • $begingroup$
    nice! I will keep the options opened in case someone comes up with an even more generic answer which would work with e.g. contour plots as well?
    $endgroup$
    – chris
    8 hours ago










  • $begingroup$
    @chris that's always a good idea, you never know what someone might come up with.
    $endgroup$
    – C. E.
    8 hours ago










  • $begingroup$
    So the next challenge would be to make it work with plots = {ContourPlot[Sin[ x y], {x, -1, 1}, {y, -1, 1}, ContourShading -> False], ContourPlot[Cos[ x - y], {x, -1, 1}, {y, -1, 1}, ContourShading -> False]}; but feel free to ignore this 'new' question.
    $endgroup$
    – chris
    7 hours ago




















  • $begingroup$
    nice! I will keep the options opened in case someone comes up with an even more generic answer which would work with e.g. contour plots as well?
    $endgroup$
    – chris
    8 hours ago










  • $begingroup$
    @chris that's always a good idea, you never know what someone might come up with.
    $endgroup$
    – C. E.
    8 hours ago










  • $begingroup$
    So the next challenge would be to make it work with plots = {ContourPlot[Sin[ x y], {x, -1, 1}, {y, -1, 1}, ContourShading -> False], ContourPlot[Cos[ x - y], {x, -1, 1}, {y, -1, 1}, ContourShading -> False]}; but feel free to ignore this 'new' question.
    $endgroup$
    – chris
    7 hours ago


















$begingroup$
nice! I will keep the options opened in case someone comes up with an even more generic answer which would work with e.g. contour plots as well?
$endgroup$
– chris
8 hours ago




$begingroup$
nice! I will keep the options opened in case someone comes up with an even more generic answer which would work with e.g. contour plots as well?
$endgroup$
– chris
8 hours ago












$begingroup$
@chris that's always a good idea, you never know what someone might come up with.
$endgroup$
– C. E.
8 hours ago




$begingroup$
@chris that's always a good idea, you never know what someone might come up with.
$endgroup$
– C. E.
8 hours ago












$begingroup$
So the next challenge would be to make it work with plots = {ContourPlot[Sin[ x y], {x, -1, 1}, {y, -1, 1}, ContourShading -> False], ContourPlot[Cos[ x - y], {x, -1, 1}, {y, -1, 1}, ContourShading -> False]}; but feel free to ignore this 'new' question.
$endgroup$
– chris
7 hours ago






$begingroup$
So the next challenge would be to make it work with plots = {ContourPlot[Sin[ x y], {x, -1, 1}, {y, -1, 1}, ContourShading -> False], ContourPlot[Cos[ x - y], {x, -1, 1}, {y, -1, 1}, ContourShading -> False]}; but feel free to ignore this 'new' question.
$endgroup$
– chris
7 hours ago




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematica Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f201318%2frecolour-existing-plots%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Hudson River Historic District Contents Geography History The district today Aesthetics Cultural...

The number designs the writing. Feandra Aversely Definition: The act of ingrafting a sprig or shoot of one...

Ayherre Geografie Demografie Externe links Navigatiemenu43° 23′ NB, 1° 15′ WL43° 23′ NB, 1°...