Skip to main content

plotting - How to 2D-plot function in two varibles


I try to make a plot of a function in multi variables as in that paper arXiv:1312.1935, FIG. 2 .


Mathematica graphics


I tried some thing like:


k[s_,f_] = s + f;

ListPlot[Table[k[s, f], {s, -1, 1, 0.5}, {f, -1, 1, 0.5}]]

But it plotted k[s, f] on the y- axis. While I'd like to have s and f on the x and y axises.



There is also ContourPlot, or PlotRegion, but to my knowledge the function k[s,f] will be plotted as continuous regions, while i'd like to present it as points (with known values like in the FIG).


So any help ?



" If the data is 3D and the third entry is obtained by applying a function like k[s,f] to the first two entries (like data set dt3d below), then the function we use to style the data is slightly different:"


Actually I don't understand from here. I understand in the example of td and styleddt that in Style[{##} and k[##], that ## refers to the two variables which k is function of them.


But now I try to plot another function, like Y[s,f,d]= s+f+d;, with -1 < s < 1, f= 0.5, and -0.5 < d < 0.5, and I want to plot Y[s,f,d] only at -2 < Y < 0, or we can use Piecewise as before to know Y values.



Answer



Update: I guess you want to color a list of 2D points using a function like k[s,f]. You can Style the original data and use the resulting data with ListPlot to get something like Figure 2 in the linked paper.


dt = RandomReal[{-1, 1}, {1000, 2}];


The first 10 rows of dt:


TableForm[dt[[;; 10]], TableHeadings -> {None, {"s", "f"}}]

Mathematica graphics


styleddt = Style[{##}, PointSize[.02],
     Piecewise[{{Orange, -1 <= k[##] <= 0}, {Green, .5 <=  k[##] <= .9}, 
      {Red, 1.2 <= k[##] <= 1.6}}, Blue]] & @@@ dt;
labels = {"-1<=k[s,f]<=0", ".5<=k[s,f]<=.9", "1.2<=k[s,f]<=1.6", "otherwise"};
colors = {Orange, Green, Red, Blue};
legend = Row[Style[##, "Panel", 18] & @@@ Transpose[{labels, colors}], Spacer[5]];


ListPlot[styleddt, DataRange -> {{-1, 1}, {-1, 1}}, Frame -> True,
 ImageSize -> 500, AspectRatio -> 1, PlotLabel -> legend]

Mathematica graphics


If the data is 3D and the third entry is obtained by applying a function like k[s,f] to the first two entries (like data set dt3d below), then the function we use to style the data is slightly different:


dt3d = {##, k@##} & @@@ dt;

The first 10 rows of dt3d:


TableForm[dt3d[[;; 10]], TableHeadings -> {None, {"s", "f", "k[s, f]"}}]


Mathematica graphics


styleddata = Style[{#, #2}, PointSize[.02],
     Piecewise[{{Orange, -1 <= #3 <=  0}, 
      {Green, .5 <= #3 <= .9}, {Red, 1.2 <= #3 <= 1.6}},  Blue]] & @@@ dt3d;

We get the same picture as above using:


ListPlot[styleddata, DataRange -> {{-1, 1}, {-1, 1}}, Frame -> True,
         ImageSize -> 500, AspectRatio -> 1, PlotLabel -> legend]


Original post:


Here are few alternative ways to use 2D plots / charts to visualize your data.


Graphics:


Graphics[{PointSize[Large],
  {ColorData[{"TemperatureMap", {-2, 2}}][#3], Disk[{#, #2}, .1], 
     Black, Text[Style[#3, 14, Bold], {#, #2}]} & @@@ (Join @@ 
     Table[{s, f, k[s, f]}, {s, -1, 1, 0.5}, {f, -1, 1, 0.5}])}, 
 Frame -> True]

Mathematica graphics



MatrixPlot


MatrixPlot[Table[k[s, f], {s, -1, 1, 0.5}, {f, -1, 1, 0.5}], 
 DataRange -> {{-1, 1}, {-1, 1}}, ColorFunction -> "TemperatureMap", 
 DataReversed -> True, PlotLegends -> Automatic]

Mathematica graphics


BubbleChart


BubbleChart[Labeled[Style[{#, #2, Abs@#3 /. 0. -> 1}, 
     ColorData[{"TemperatureMap", {-2, 2}}][#3]], #3] & @@@ (Join @@ 
    Table[{s, f, k[s, f]}, {s, -1, 1, 0.5}, {f, -1, 1, 0.5}]), 

 BubbleScale -> (1 &)]

Mathematica graphics


Comments

Post a Comment

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}]
Post a Comment
Read more

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}]
Post a Comment
Read more

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....
Post a Comment
Read more