Skip to main content

plotting - ListLinePlot with vertical filling gradient


I am aware that the topic of vertical filling has been discussed before (e.g., here and here), but I could not find any reasonably simple solution to the following problem:


I have a discrete data set which I plot with ListLinePlot. I want to fill the area between the data and a given y value if the data is below that value. That much is easy:


data = Table[{x, (x - 5)^2 + 3}, {x, 0, 10}];

Show[ListLinePlot[data,

Filling -> {1 -> {15, {Directive[Blue, Opacity[0.1]], None}}}],
ListPlot[data]]

enter image description here


However, now I want the filling to have a vertical gradient in opacity (or white level, I don't really mind) such that the filling is blue at y=3 and transparent for y>15. I found a solution for filling under all data points using polygons and for parametrizable functions. However, as it happens, my filling does not end at a data point and it seems to me that applying the polygon solution would be quite a hack that involves first finding and constructing proper boundary points for the polygon.


Is there really no simpler way for vertical gradients?


Edit:


Just for the sake of completeness, I realized that the above link (filling under all data points) does give a suggestion of how to use ParametricPlot together with Interpolate to create a vertical gradient filling on discrete data. The code would look like this:


interp = Interpolation[data, InterpolationOrder -> 1];
Show[ListLinePlot[data],

ParametricPlot[{x,
With[{value = 15 (1 - y) + interp[x] y},
If[value > interp[x], value]]}, {x, 0, 10}, {y, 0, 1},
ColorFunction -> (Blend[{Blue, White}, #2] &),
BoundaryStyle -> None]]

However, the result is much less appealing then the polygon solution. It is possible to improve the quality by ramping up the PlotPoints, but then it takes very long to calculate the graph and exporting it, in particular to pdf, crashes Mathematica completely on my computer.


enter image description here



Answer



As you say, it's a hack, but it's a relatively small one:



p1 = ListLinePlot[data, Filling -> {1 -> {15, {Blue, None}}}, PlotMarkers -> Automatic]

(I changed the Filling parameters on a suggestion by the OP.)


Then:


Normal@p1 /. Polygon[x_] :> Polygon[x, VertexColors -> (Blend[{Blue, White}, #] & /@ Rescale[x[[All, 2]]])]

enter image description here


The Rescale stuff is meant to re-scale the y-values so that they go between 0 and 1, allowing us to Blend from Blue to White. (Thanks to MarcoB for the nicer code.)


Comments

Popular posts from this blog

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 - 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 - 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....