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Reverse ColorFunction or ColorData
ListContourPlot-ColorFunctionScaled ColorData FunctionPlotting: DensityPlot produces artifactsParametricPlot - Problem with ColorFunctionDensityPlot: giving the same color to a range of valuesSpecify range and reverse for colordata in contourplotrescaling ColorFunction for ListContourPlotAdding a second color bar to a plotColorFunction in GraphicsColorData input arguments not clear
$begingroup$
please excuse my bad English, I'm not a native speaker.
This should be fairly simple: Is there a simple way to "reverse" ColorFunction or ColorData?
Let me explain what I mean. Let's say, we have the following plot:
Plot[x, x, 1, 10, ColorFunction -> "Rainbow"]
As you can see, the upper part of the plot is colored red, the lower part is colored blue/purple. How do I reverse that? So that the upper part is purple, the lower part red.
Thank you for your help. I appreciate it.
If you're curious why I ask that: I have made this plot showing the number of molecules of three biological substances over time when metabolized by the body.
plotting graphics color
asked 3 hours ago
Marian StiehlerMarian Stiehler
855
$endgroup$
add a comment |
$begingroup$
please excuse my bad English, I'm not a native speaker.
This should be fairly simple: Is there a simple way to "reverse" ColorFunction or ColorData?
Let me explain what I mean. Let's say, we have the following plot:
Plot[x, x, 1, 10, ColorFunction -> "Rainbow"]
As you can see, the upper part of the plot is colored red, the lower part is colored blue/purple. How do I reverse that? So that the upper part is purple, the lower part red.
Thank you for your help. I appreciate it.
If you're curious why I ask that: I have made this plot showing the number of molecules of three biological substances over time when metabolized by the body.
plotting graphics color
asked 3 hours ago
Marian StiehlerMarian Stiehler
855
$endgroup$
add a comment |
$begingroup$
please excuse my bad English, I'm not a native speaker.
This should be fairly simple: Is there a simple way to "reverse" ColorFunction or ColorData?
Let me explain what I mean. Let's say, we have the following plot:
Plot[x, x, 1, 10, ColorFunction -> "Rainbow"]
As you can see, the upper part of the plot is colored red, the lower part is colored blue/purple. How do I reverse that? So that the upper part is purple, the lower part red.
Thank you for your help. I appreciate it.
If you're curious why I ask that: I have made this plot showing the number of molecules of three biological substances over time when metabolized by the body.
plotting graphics color
asked 3 hours ago
Marian StiehlerMarian Stiehler
855
$endgroup$
please excuse my bad English, I'm not a native speaker.
This should be fairly simple: Is there a simple way to "reverse" ColorFunction or ColorData?
Let me explain what I mean. Let's say, we have the following plot:
Plot[x, x, 1, 10, ColorFunction -> "Rainbow"]
As you can see, the upper part of the plot is colored red, the lower part is colored blue/purple. How do I reverse that? So that the upper part is purple, the lower part red.
Thank you for your help. I appreciate it.
If you're curious why I ask that: I have made this plot showing the number of molecules of three biological substances over time when metabolized by the body.
plotting graphics color
plotting graphics color
asked 3 hours ago
Marian StiehlerMarian Stiehler
855
asked 3 hours ago
Marian StiehlerMarian Stiehler
855
asked 3 hours ago
Marian StiehlerMarian Stiehler
855
asked 3 hours ago
Marian StiehlerMarian Stiehler
855
asked 3 hours ago
Marian StiehlerMarian Stiehler
855
855
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
This may be a shorter solution.
Plot[x, x, 1, 10, ColorFunction -> ColorData["Rainbow", "Reverse"]]
answered 3 hours ago
L.YuL.Yu
886
New contributor
L.Yu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
To reverse the colours you could try 1-# instead of just # in a pure function passed to ColorFunction. To change how the line segments are coloured, include an exponent as follows.
Manipulate[
Plot[E^-x, -1.2 x, -2.4 x, x, 0, 5,
ColorFunction -> (ColorData["Rainbow", (1 - #)^e] &),
PlotRange -> 0, 5, 0, 1, Frame -> True, ImageSize -> 450,
FrameLabel -> "Zeit in Minuten", "Anzahl der Wirkstoffmoleküle"],
e, 1.0, "Exponent", 0, 10, Appearance -> "Labeled"]
answered 2 hours ago
KennyColnagoKennyColnago
12.3k1754
$endgroup$
add a comment |
$begingroup$
I have found one possible solution… :)
Plot[x, x, 1, 10, ColorFunction -> Function[x, y, Reverse@ColorData["Rainbow"][y]]]
Is there a better way?
answered 3 hours ago
Marian StiehlerMarian Stiehler
855
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This may be a shorter solution.
Plot[x, x, 1, 10, ColorFunction -> ColorData["Rainbow", "Reverse"]]
answered 3 hours ago
L.YuL.Yu
886
New contributor
L.Yu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
This may be a shorter solution.
Plot[x, x, 1, 10, ColorFunction -> ColorData["Rainbow", "Reverse"]]
answered 3 hours ago
L.YuL.Yu
886
New contributor
L.Yu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
This may be a shorter solution.
Plot[x, x, 1, 10, ColorFunction -> ColorData["Rainbow", "Reverse"]]
answered 3 hours ago
L.YuL.Yu
886
New contributor
L.Yu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
This may be a shorter solution.
Plot[x, x, 1, 10, ColorFunction -> ColorData["Rainbow", "Reverse"]]
answered 3 hours ago
L.YuL.Yu
886
New contributor
L.Yu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
L.YuL.Yu
886
New contributor
L.Yu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 3 hours ago
L.YuL.Yu
886
answered 3 hours ago
L.YuL.Yu
886
886
New contributor
L.Yu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
L.Yu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
L.Yu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
To reverse the colours you could try 1-# instead of just # in a pure function passed to ColorFunction. To change how the line segments are coloured, include an exponent as follows.
Manipulate[
Plot[E^-x, -1.2 x, -2.4 x, x, 0, 5,
ColorFunction -> (ColorData["Rainbow", (1 - #)^e] &),
PlotRange -> 0, 5, 0, 1, Frame -> True, ImageSize -> 450,
FrameLabel -> "Zeit in Minuten", "Anzahl der Wirkstoffmoleküle"],
e, 1.0, "Exponent", 0, 10, Appearance -> "Labeled"]
answered 2 hours ago
KennyColnagoKennyColnago
12.3k1754
$endgroup$
add a comment |
$begingroup$
To reverse the colours you could try 1-# instead of just # in a pure function passed to ColorFunction. To change how the line segments are coloured, include an exponent as follows.
Manipulate[
Plot[E^-x, -1.2 x, -2.4 x, x, 0, 5,
ColorFunction -> (ColorData["Rainbow", (1 - #)^e] &),
PlotRange -> 0, 5, 0, 1, Frame -> True, ImageSize -> 450,
FrameLabel -> "Zeit in Minuten", "Anzahl der Wirkstoffmoleküle"],
e, 1.0, "Exponent", 0, 10, Appearance -> "Labeled"]
answered 2 hours ago
KennyColnagoKennyColnago
12.3k1754
$endgroup$
add a comment |
$begingroup$
To reverse the colours you could try 1-# instead of just # in a pure function passed to ColorFunction. To change how the line segments are coloured, include an exponent as follows.
Manipulate[
Plot[E^-x, -1.2 x, -2.4 x, x, 0, 5,
ColorFunction -> (ColorData["Rainbow", (1 - #)^e] &),
PlotRange -> 0, 5, 0, 1, Frame -> True, ImageSize -> 450,
FrameLabel -> "Zeit in Minuten", "Anzahl der Wirkstoffmoleküle"],
e, 1.0, "Exponent", 0, 10, Appearance -> "Labeled"]
answered 2 hours ago
KennyColnagoKennyColnago
12.3k1754
$endgroup$
To reverse the colours you could try 1-# instead of just # in a pure function passed to ColorFunction. To change how the line segments are coloured, include an exponent as follows.
Manipulate[
Plot[E^-x, -1.2 x, -2.4 x, x, 0, 5,
ColorFunction -> (ColorData["Rainbow", (1 - #)^e] &),
PlotRange -> 0, 5, 0, 1, Frame -> True, ImageSize -> 450,
FrameLabel -> "Zeit in Minuten", "Anzahl der Wirkstoffmoleküle"],
e, 1.0, "Exponent", 0, 10, Appearance -> "Labeled"]
answered 2 hours ago
KennyColnagoKennyColnago
12.3k1754
answered 2 hours ago
KennyColnagoKennyColnago
12.3k1754
answered 2 hours ago
KennyColnagoKennyColnago
12.3k1754
answered 2 hours ago
KennyColnagoKennyColnago
12.3k1754
12.3k1754
add a comment |
add a comment |
$begingroup$
I have found one possible solution… :)
Plot[x, x, 1, 10, ColorFunction -> Function[x, y, Reverse@ColorData["Rainbow"][y]]]
Is there a better way?
answered 3 hours ago
Marian StiehlerMarian Stiehler
855
$endgroup$
add a comment |
$begingroup$
I have found one possible solution… :)
Plot[x, x, 1, 10, ColorFunction -> Function[x, y, Reverse@ColorData["Rainbow"][y]]]
Is there a better way?
answered 3 hours ago
Marian StiehlerMarian Stiehler
855
$endgroup$
add a comment |
$begingroup$
I have found one possible solution… :)
Plot[x, x, 1, 10, ColorFunction -> Function[x, y, Reverse@ColorData["Rainbow"][y]]]
Is there a better way?
answered 3 hours ago
Marian StiehlerMarian Stiehler
855
$endgroup$
I have found one possible solution… :)
Plot[x, x, 1, 10, ColorFunction -> Function[x, y, Reverse@ColorData["Rainbow"][y]]]
Is there a better way?
answered 3 hours ago
Marian StiehlerMarian Stiehler
855
answered 3 hours ago
Marian StiehlerMarian Stiehler
855
answered 3 hours ago
Marian StiehlerMarian Stiehler
855
answered 3 hours ago
Marian StiehlerMarian Stiehler
855
855
add a comment |
add a comment |
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