Skip to main content

bugs - How can I change the thickness of tick marks in BarLegend?


Bug introduced in 10.1 or earlier and persisting through 11.0.1 or later



This answer explains how to change various undocumented options for BarLegend with Method. In particular, I want to change the style of the ticks (marks and labels) in BarLegend. For example,


BarLegend[{"SunsetColors", {0, 1}}, LabelStyle -> {FontSize -> 12},

 Method -> {Frame -> False, TicksStyle -> Directive[Red, AbsoluteThickness[2]]}]

However, in Mathematica 10.2 on Linux, BarLegend refuses to change the thickness of the tick marks. Also, if I do not add the LabelStyle option then the labels don't change color? In Plot this command works,


Plot[Sin[x], {x, 0, 2 Pi},TicksStyle -> Directive[Red, AbsoluteThickness[2]]]

so I would expect that it works for BarLegend as well.


What is going on and is there a workaround to change the thickness of the tick marks? Thanks.



Answer



Somehow the AbsolutThickness you specified gets replaced by a default value of AbsoluteThickness[0.2].


This misbehavior can be corrected by replacing the incorrect value with your specification.



PlotLegends; (*preload definitions*)

Cell[BoxData[
   MakeBoxes@
      BarLegend[{"SunsetColors", {0, 1}}, LabelStyle -> {FontSize -> 12}, 
       Method -> {Frame -> False, TicksStyle -> Directive[Red, AbsoluteThickness[2]]}] /. 
     Directive[RGBColor[1, 0, 0], AbsoluteThickness[_]] -> 
      Directive[RGBColor[1, 0, 0], AbsoluteThickness[2]] // #[[1, 1]] &
   ], "Output"] // CellPrint


AbsoluteThickness[2]


For opaque ticks:


Cell[BoxData[
   MakeBoxes@
      BarLegend[{"SunsetColors", {0, 1}}, LabelStyle -> {FontSize -> 12}, 
       Method -> {Frame -> False, TicksStyle -> Directive[Red, AbsoluteThickness[2]]}] /. 
     Directive[RGBColor[1, 0, 0], AbsoluteThickness[_]] -> 
      Directive[RGBColor[1, 0, 0], AbsoluteThickness[2], Opacity[1]] // #[[1, 1]] &
   ], "Output"] // CellPrint


opaque



Correcting the BarLegend of a DensityPlot, using the syntax provided in the answer by Praan :


DensityPlot[Sin[x y], {x, 0, 1}, {y, 0, 1}, 
  PlotLegends -> 
   BarLegend[Automatic, LabelStyle -> {FontSize -> 12}, 
    Method -> {Frame -> False, TicksStyle -> Directive[Red, AbsoluteThickness[2]]}]] /. 
 Placed[barLegend_BarLegend, args__] :> 
  Placed[ToExpression[
    FrameBox @@ MakeBoxes[barLegend] /. 

      Directive[Red, AbsoluteThickness[_]] -> 
       Directive[Red, AbsoluteThickness[2], Opacity[1]]], args]

DensityPlot


The same output can be achieved by using the following LegendFunction


DensityPlot[Sin[x y], {x, 0, 1}, {y, 0, 1}, 
 PlotLegends -> 
  BarLegend[Automatic, LabelStyle -> {FontSize -> 12}, 
   Method -> {Frame -> False, 
     TicksStyle -> Directive[Red, AbsoluteThickness[2]]}, 

   LegendFunction -> (# /. 
       Directive[Red, AbsoluteThickness[_]] -> 
        Directive[Red, AbsoluteThickness[2], Opacity[1]] &)]]


With the answer by Praan and our discussion in the comments it became clear, that a wrong InterpretationFunction inside the TemplateBox created by BarLegend can cause additional problems.
Compare


MakeBoxes[
    BarLegend[{"SunsetColors", {0, 1}}, LegendMarkerSize -> 300, 
     LabelStyle -> {FontSize -> 12}, 

     Method -> {FrameStyle -> Black, AxesStyle -> None, 
       TicksStyle -> Black}]] /. 
   AbsoluteThickness[_] -> 
    AbsoluteThickness[2] /. (InterpretationFunction :> 
     f_) -> (InterpretationFunction :> (# &)) // ToExpression

corr


with


MakeBoxes[
   BarLegend[{"SunsetColors", {0, 1}}, LegendMarkerSize -> 300, 

    LabelStyle -> {FontSize -> 12}, 
    Method -> {FrameStyle -> Black, AxesStyle -> None, 
      TicksStyle -> Black}]] /. 
  AbsoluteThickness[_] -> AbsoluteThickness[2] // ToExpression

false


or just the InterpretationFunction


MakeBoxes[
    BarLegend[{"SunsetColors", {0, 1}}, LegendMarkerSize -> 300, 
     LabelStyle -> {FontSize -> 12}, 

     Method -> {FrameStyle -> Black, AxesStyle -> None, 
       TicksStyle -> Directive[Black, AbsoluteThickness[2]]}]] /. 
   AbsoluteThickness[_] -> 
    AbsoluteThickness[2] // #[[-1, 2, 1]] & // ToExpression

and the first code block in the answer by Praan.


Comments

Post a Comment

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}]
Post a Comment
Read more

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 - 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