Skip to main content

charts - Retrieving PlotRange from BarChart


Consider the following data:


data1={73.9377, 54.4122, 53.0826, 24.1936};
data2={76.828, 49.1673, 45.7883, 18.9015};

I defined my own BarChart as follows:


MyBarCharts[list_]:= 
BarChart[
list,

Frame -> {{True, None}, {True, None}}
]

When applying MyBarCharts to data1 and data2 I get (not really surprising) two different (with regard to the coordinate system) plots. Now I would like to have for both plots the exact same coordinate system. I know that with Histogram it works as follows:


Histogram[
data2,
PlotRange -> Part[PlotRange /. Options[Histogram[
data1,
PlotRange -> Automatic
],PlotRange], 2],

Frame -> {{True, None}, {True, None}}]

But the implementation of that approach in MyBarCharts does not give the desired result. Has anyone an idea?



Answer



Extracting the PlotRange from a BarChart is not as straightforward as it should be. If no PlotRange is specified in creating the chart, then Options will return PlotRange -> All and AbsoluteOptions will return PlotRange -> {{0., 1.}, {0., 1.}


bc = BarChart[{1, 2, 3, 4}];

Options[bc, PlotRange]
(* {PlotRange -> All} *)


AbsoluteOptions[bc, PlotRange]
(* {PlotRange -> {{0., 1.}, {0., 1.}}} *)

The incorrect result from AbsoluteOptions appears to be related to the presence of dynamic objects in the graphics expression (i.e. the bars with their mouseover effects) . I don't know why this causes AbsoluteOptions to go wrong, but a workaround is to replace the dynamic bars with straighforward rectangles, allowing AbsoluteOptions to extract the correct PlotRange. Thus, a replacement for AbsoluteOptions for BarCharts is:


barChartOptions[chart_, opts___] := 
AbsoluteOptions[chart /.
Tooltip[StatusArea[RawBoxes[DynamicBox[{_, RectangleBox[data__]}]], _], _] :>
Rectangle[data], opts]

barChartOptions[bc, PlotRange]

(* {PlotRange -> {{0.545455, 4.45455}, {0., 4.}}} *)

Comments

Popular posts from this blog

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.