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

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
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