plotting - Getting a correct custom legend for contourplot?

I am having trouble making a custom range for a legend in ContourPlot since the legend doesn't "talk" to the PlotRange and adjust its scale accordingly. How do I go about fixing this?

As an explicit example consider the following. If I run ContourPlot with an automatic legend I get the right scaling:

 ContourPlot[x y , {x, 0, 2}, {y, 0, 2}, PlotLegends->Automatic,

ColorFunction ->"BlueGreenYellow"]

enter image description here

But if I use a custom legend range then the legend no longer corresponds to the image:

ContourPlot[x y , {x, 0, 2}, {y, 0, 2},
PlotLegends -> BarLegend[{"BlueGreenYellow", {0, 10}}]

enter image description here

How do I get the plot to use the same scale as the legend?

Answer

This was more involved than I expected. First, you need to set the ColorFunction to encompass the full range,

ColorData[{"BlueGreenYellow", {0, 10}}]

Interestingly, I did not know about that form of ColorData until earlier this week, so I recommend reading through the Details section closely. Now, to use this within ContourPlot, you need to set

ColorFunctionScaling -> False

otherwise ContourPlot will rescale everything. As to the legend, you need to specify both the contour range, like you were doing, and the number of contours to force it to use the range correctly,

BarLegend[{Automatic, {0, 10}}, 20]

Note the use of Automatic, the color function is automatically passed to the legend, so you do not to add it again. The range is necessary because the legend is only passed the range present in the plot itself. Lastly, the number of contours has to be large enough for BarLegend to use the full range, and 20 works here. All together, this is

ContourPlot[x y, {x, 0, 2}, {y, 0, 2},
  PlotLegends -> BarLegend[{Automatic, {0, 10}}, 20], 
  ColorFunction -> ColorData[{"BlueGreenYellow", {0, 10}}],
  ColorFunctionScaling -> False
]

enter image description here

Comments

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.

Blog

Search This Blog