Skip to main content

plotting - Extracting the coordinate of a particular point of interest from a ListPlot


Is there a way to obtain the coordinate of a point of interest in a ListPlot?


As an example, I have a list containing many sets of 2D coordinates and the plot drawn is discontinuous at one point (the first derivative is not continuous and the gradient increases suddenly).


Can I extract the location of that point interactively? Otherwise, I have to search through the list of data myself to determine the change of gradient, which defeats the whole purpose of drawing a plot. Also, using the Get Coordinates function from the right click menu does not give very accurate results.



Answer



ListPlot accepts data wrappers besides Tooltip


(although I could not find any mention of this feature in the docs).


So, @Jens' method can be achieved without post-processing:


 data = Table[{Sin[n], Sin[2 n]}, {n, 50}];
ListPlot[PopupWindow[Tooltip[#], #] & /@ data]


enter image description here


On mouseover:


enter image description here


Click on a point:


enter image description here


Note: Thought this was a new feature added in Version-9, but as @Alexey Popkov noted it also works in version 8.0.4, so it has been around for some time.


Update: A simpler version of @Mr.Wizard's printTip can also be used as a wrapper directly inside ListPlot:


 ListPlot[Button[Tooltip@#, Print[#]] & /@ N@data]


enter image description here


Update 2: Collecting point coordinates:


 clicks = {};
Column[{ListPlot[Button[Tooltip@#, AppendTo[clicks, #]] & /@ N@data,
ImageSize -> 300],
"\n\t", Row[{"clicks = " , Dynamic[clicks // TableForm]}]}]

enter image description here


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....