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ColorFunction based on array index in ListLinePlot

ListContourPlot-ColorFunctionUse ColorFunction in ListLinePlot with IfAdapt ColorFunction in Array PlotColorFunction based on dataColorFunction for even rows in an arrayWhy does not ListLinePlot work with a new defined ColorFunction?Independent Colorfunction in RevolutionPlot3D

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

3

$begingroup$

ColorFunction specifies that for ListLinePlot it takes the $x,y$ data as inputs. I, however, would like to take the array index (or in general some external array of the same length) as the input for ColorFunction so that, if we use a rainbow color scheme, the earlier points show up as purple while the last points show up as red. An application would be visualisation of the long time behaviour of some system when data is imported from an external source.

For example if we take

ListLinePlot[Table[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 0.1]]

then the outer lines should be purple ending up as red as the plot spirals in.

plotting

share|improve this question

edited 8 hours ago

Takoda

asked 8 hours ago

TakodaTakoda

36188 bronze badges

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add a comment
 | 

3

$begingroup$

ColorFunction specifies that for ListLinePlot it takes the $x,y$ data as inputs. I, however, would like to take the array index (or in general some external array of the same length) as the input for ColorFunction so that, if we use a rainbow color scheme, the earlier points show up as purple while the last points show up as red. An application would be visualisation of the long time behaviour of some system when data is imported from an external source.

For example if we take

ListLinePlot[Table[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 0.1]]

then the outer lines should be purple ending up as red as the plot spirals in.

plotting

share|improve this question

edited 8 hours ago

Takoda

asked 8 hours ago

TakodaTakoda

36188 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

ColorFunction specifies that for ListLinePlot it takes the $x,y$ data as inputs. I, however, would like to take the array index (or in general some external array of the same length) as the input for ColorFunction so that, if we use a rainbow color scheme, the earlier points show up as purple while the last points show up as red. An application would be visualisation of the long time behaviour of some system when data is imported from an external source.

For example if we take

ListLinePlot[Table[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 0.1]]

then the outer lines should be purple ending up as red as the plot spirals in.

plotting

share|improve this question

edited 8 hours ago

Takoda

asked 8 hours ago

TakodaTakoda

36188 bronze badges

$endgroup$

ListLinePlot[Table[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 0.1]]

share|improve this question

edited 8 hours ago

Takoda

asked 8 hours ago

TakodaTakoda

36188 bronze badges

share|improve this question

edited 8 hours ago

Takoda

asked 8 hours ago

TakodaTakoda

36188 bronze badges

asked 8 hours ago

TakodaTakoda

36188 bronze badges

asked 8 hours ago

TakodaTakoda

36188 bronze badges

3 Answers
3

active

oldest

votes

2

$begingroup$

data = Table[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 0.1];

indexlist = Rescale[Range @ Length @ data] (* or your external list to control color*);

iF = Interpolation[MapThread[#, #2 &, data, indexlist], InterpolationOrder -> 1];

ListLinePlot[data,
 AspectRatio -> 1, 
 ColorFunctionScaling -> False, 
 ColorFunction -> (ColorData["Rainbow", "Reversed"][ iF[#, #2]]&)]

Alternative methods:

PolarPlot

PolarPlot[E^-0.001 x , x, 0, 100, 
 ColorFunction -> (ColorData["Rainbow", "Reversed"][#3] &)

same picture

ParametricPlot

ParametricPlot[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 
 ColorFunction -> (ColorData["Rainbow", "Reversed"][#3] &)]

same picture

Graphics + VertexColors

Graphics[Line[data, VertexColors -> (ColorData["Rainbow", "Reversed"] /@ indexlist)], 
 Axes -> True]

same picture

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

216k1010 gold badges246246 silver badges496496 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a good answer, but I think you want (ColorData["Rainbow", "Reverse"] to satisfy the OP's coloring order.
  $endgroup$
  – m_goldberg
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you @m_goldberg. ColorData["Rainbow", "Reversed"] and ColorData["Rainbow", "Reverse"] both work.
  $endgroup$
  – kglr
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, they both work. Your solution is independent of the ordering of the Rainbow color scheme. Still, I feel a little stupid about making my original comment.
  $endgroup$
  – m_goldberg
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

1

$begingroup$

With[lines=Line/@Partition[Table[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 0.1],2,1], 
Graphics[MapIndexed[ColorData["Rainbow","Reverse"][First@#2/Length[lines]],#1&,lines],
Frame -> True]]

share|improve this answer

answered 7 hours ago

AlxAlx

88844 silver badges99 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

I think it is easier to generate the plot you want with ParametricPlot Like so:

With[n = 100,
 ParametricPlot[E^-0.001 x Cos[x], Sin[x], x, 0, n,
 Mesh -> n - 1,
 MeshStyle -> Transparent,
 MeshShading ->Table[ColorData[L"Rainbow", "Reverse"][i/n], i, n]]]

share|improve this answer

edited 7 hours ago

answered 7 hours ago

m_goldbergm_goldberg

91.6k88 gold badges7575 silver badges209209 bronze badges

$endgroup$

add a comment
 | 

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2

$begingroup$

data = Table[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 0.1];

indexlist = Rescale[Range @ Length @ data] (* or your external list to control color*);

iF = Interpolation[MapThread[#, #2 &, data, indexlist], InterpolationOrder -> 1];

ListLinePlot[data,
 AspectRatio -> 1, 
 ColorFunctionScaling -> False, 
 ColorFunction -> (ColorData["Rainbow", "Reversed"][ iF[#, #2]]&)]

Alternative methods:

PolarPlot

PolarPlot[E^-0.001 x , x, 0, 100, 
 ColorFunction -> (ColorData["Rainbow", "Reversed"][#3] &)

same picture

ParametricPlot

ParametricPlot[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 
 ColorFunction -> (ColorData["Rainbow", "Reversed"][#3] &)]

same picture

Graphics + VertexColors

Graphics[Line[data, VertexColors -> (ColorData["Rainbow", "Reversed"] /@ indexlist)], 
 Axes -> True]

same picture

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

216k1010 gold badges246246 silver badges496496 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a good answer, but I think you want (ColorData["Rainbow", "Reverse"] to satisfy the OP's coloring order.
  $endgroup$
  – m_goldberg
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you @m_goldberg. ColorData["Rainbow", "Reversed"] and ColorData["Rainbow", "Reverse"] both work.
  $endgroup$
  – kglr
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, they both work. Your solution is independent of the ordering of the Rainbow color scheme. Still, I feel a little stupid about making my original comment.
  $endgroup$
  – m_goldberg
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

2

$begingroup$

data = Table[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 0.1];

indexlist = Rescale[Range @ Length @ data] (* or your external list to control color*);

iF = Interpolation[MapThread[#, #2 &, data, indexlist], InterpolationOrder -> 1];

ListLinePlot[data,
 AspectRatio -> 1, 
 ColorFunctionScaling -> False, 
 ColorFunction -> (ColorData["Rainbow", "Reversed"][ iF[#, #2]]&)]

Alternative methods:

PolarPlot

PolarPlot[E^-0.001 x , x, 0, 100, 
 ColorFunction -> (ColorData["Rainbow", "Reversed"][#3] &)

same picture

ParametricPlot

ParametricPlot[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 
 ColorFunction -> (ColorData["Rainbow", "Reversed"][#3] &)]

same picture

Graphics + VertexColors

Graphics[Line[data, VertexColors -> (ColorData["Rainbow", "Reversed"] /@ indexlist)], 
 Axes -> True]

same picture

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

216k1010 gold badges246246 silver badges496496 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a good answer, but I think you want (ColorData["Rainbow", "Reverse"] to satisfy the OP's coloring order.
  $endgroup$
  – m_goldberg
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you @m_goldberg. ColorData["Rainbow", "Reversed"] and ColorData["Rainbow", "Reverse"] both work.
  $endgroup$
  – kglr
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right, they both work. Your solution is independent of the ordering of the Rainbow color scheme. Still, I feel a little stupid about making my original comment.
  $endgroup$
  – m_goldberg
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

$begingroup$

data = Table[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 0.1];

indexlist = Rescale[Range @ Length @ data] (* or your external list to control color*);

iF = Interpolation[MapThread[#, #2 &, data, indexlist], InterpolationOrder -> 1];

ListLinePlot[data,
 AspectRatio -> 1, 
 ColorFunctionScaling -> False, 
 ColorFunction -> (ColorData["Rainbow", "Reversed"][ iF[#, #2]]&)]

Alternative methods:

PolarPlot

PolarPlot[E^-0.001 x , x, 0, 100, 
 ColorFunction -> (ColorData["Rainbow", "Reversed"][#3] &)

same picture

ParametricPlot

ParametricPlot[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 
 ColorFunction -> (ColorData["Rainbow", "Reversed"][#3] &)]

same picture

Graphics + VertexColors

Graphics[Line[data, VertexColors -> (ColorData["Rainbow", "Reversed"] /@ indexlist)], 
 Axes -> True]

same picture

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

216k1010 gold badges246246 silver badges496496 bronze badges

$endgroup$

data = Table[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 0.1];

indexlist = Rescale[Range @ Length @ data] (* or your external list to control color*);

iF = Interpolation[MapThread[#, #2 &, data, indexlist], InterpolationOrder -> 1];

ListLinePlot[data,
 AspectRatio -> 1, 
 ColorFunctionScaling -> False, 
 ColorFunction -> (ColorData["Rainbow", "Reversed"][ iF[#, #2]]&)]

PolarPlot[E^-0.001 x , x, 0, 100, 
 ColorFunction -> (ColorData["Rainbow", "Reversed"][#3] &)

ParametricPlot[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 
 ColorFunction -> (ColorData["Rainbow", "Reversed"][#3] &)]

Graphics[Line[data, VertexColors -> (ColorData["Rainbow", "Reversed"] /@ indexlist)], 
 Axes -> True]

share|improve this answer

edited 7 hours ago

answered 8 hours ago

kglrkglr

216k1010 gold badges246246 silver badges496496 bronze badges

answered 8 hours ago

kglrkglr

216k1010 gold badges246246 silver badges496496 bronze badges

answered 8 hours ago

kglrkglr

216k1010 gold badges246246 silver badges496496 bronze badges

1

$begingroup$

With[lines=Line/@Partition[Table[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 0.1],2,1], 
Graphics[MapIndexed[ColorData["Rainbow","Reverse"][First@#2/Length[lines]],#1&,lines],
Frame -> True]]

share|improve this answer

answered 7 hours ago

AlxAlx

88844 silver badges99 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

With[lines=Line/@Partition[Table[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 0.1],2,1], 
Graphics[MapIndexed[ColorData["Rainbow","Reverse"][First@#2/Length[lines]],#1&,lines],
Frame -> True]]

share|improve this answer

answered 7 hours ago

AlxAlx

88844 silver badges99 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

With[lines=Line/@Partition[Table[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 0.1],2,1], 
Graphics[MapIndexed[ColorData["Rainbow","Reverse"][First@#2/Length[lines]],#1&,lines],
Frame -> True]]

share|improve this answer

answered 7 hours ago

AlxAlx

88844 silver badges99 bronze badges

$endgroup$

With[lines=Line/@Partition[Table[E^-0.001 x Cos[x], Sin[x], x, 0, 100, 0.1],2,1], 
Graphics[MapIndexed[ColorData["Rainbow","Reverse"][First@#2/Length[lines]],#1&,lines],
Frame -> True]]

share|improve this answer

answered 7 hours ago

AlxAlx

88844 silver badges99 bronze badges

answered 7 hours ago

AlxAlx

88844 silver badges99 bronze badges

answered 7 hours ago

AlxAlx

88844 silver badges99 bronze badges

1

$begingroup$

I think it is easier to generate the plot you want with ParametricPlot Like so:

With[n = 100,
 ParametricPlot[E^-0.001 x Cos[x], Sin[x], x, 0, n,
 Mesh -> n - 1,
 MeshStyle -> Transparent,
 MeshShading ->Table[ColorData[L"Rainbow", "Reverse"][i/n], i, n]]]

share|improve this answer

edited 7 hours ago

answered 7 hours ago

m_goldbergm_goldberg

91.6k88 gold badges7575 silver badges209209 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

I think it is easier to generate the plot you want with ParametricPlot Like so:

With[n = 100,
 ParametricPlot[E^-0.001 x Cos[x], Sin[x], x, 0, n,
 Mesh -> n - 1,
 MeshStyle -> Transparent,
 MeshShading ->Table[ColorData[L"Rainbow", "Reverse"][i/n], i, n]]]

share|improve this answer

edited 7 hours ago

answered 7 hours ago

m_goldbergm_goldberg

91.6k88 gold badges7575 silver badges209209 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

I think it is easier to generate the plot you want with ParametricPlot Like so:

With[n = 100,
 ParametricPlot[E^-0.001 x Cos[x], Sin[x], x, 0, n,
 Mesh -> n - 1,
 MeshStyle -> Transparent,
 MeshShading ->Table[ColorData[L"Rainbow", "Reverse"][i/n], i, n]]]

share|improve this answer

edited 7 hours ago

answered 7 hours ago

m_goldbergm_goldberg

91.6k88 gold badges7575 silver badges209209 bronze badges

$endgroup$

With[n = 100,
 ParametricPlot[E^-0.001 x Cos[x], Sin[x], x, 0, n,
 Mesh -> n - 1,
 MeshStyle -> Transparent,
 MeshShading ->Table[ColorData[L"Rainbow", "Reverse"][i/n], i, n]]]

share|improve this answer

edited 7 hours ago

answered 7 hours ago

m_goldbergm_goldberg

91.6k88 gold badges7575 silver badges209209 bronze badges

answered 7 hours ago

m_goldbergm_goldberg

91.6k88 gold badges7575 silver badges209209 bronze badges

answered 7 hours ago

m_goldbergm_goldberg

91.6k88 gold badges7575 silver badges209209 bronze badges