Skip to main content

output formatting - Using Padding with Text and Style


Suppose that I use:


Text[Style["(1, 0, 0)", 12, Background -> White], {1, 1, 0}]

How can I put a few points of padding around the text?



Thanks to Jens:


Initialization cell:


Options[myText] = {Background -> White, Padding -> 1, FrameStyle -> None};


myText[s_, pos_, OptionsPattern[]] := 
 Text[Framed[Style[s, 12], Background -> OptionValue[Background], 
   FrameMargins -> OptionValue[Padding], 
   FrameStyle -> OptionValue[FrameStyle]], pos]

Image code using myText:


Show[

 ContourPlot3D[{z == Sqrt[x^2 + y^2]}, {x, -2.5, 2.5}, {y, -2.5, 
   2.5}, {z, 0, 2},
  ContourStyle -> {Yellow, Opacity[0.8]}, Mesh -> None],
 ParametricPlot3D[{r Cos[t], r Sin[t], 2}, {t, 0, 2 π}, {r, 0, 2},
   PlotStyle -> {Blue, Opacity[0.5]}, Mesh -> None],
 ParametricPlot3D[{r Cos[t], r Sin[t], 0}, {t, 0, 2 π}, {r, 0, 2},
   PlotStyle -> {LightBlue, Opacity[0.5]}, Mesh -> None],
 Graphics3D[{
   Red, Thick, Arrow[{{0, 0, 0}, {2, 2, 0}}],
   Black, PointSize[Large],

   Point[{{0, 0, 0}, 2 {Cos[Pi/4], Sin[Pi/4], 0}}],
   myText["r = 0", {0, -0.6, 0}],
   myText["r = 2", 2 {Cos[30 °], Sin[30 °], 0}],
   Red, Thick, 
   Arrow[{{Cos[Pi/4], Sin[Pi/4], 0}, {Cos[Pi/4], Sin[Pi/4], 2.5}}],
   Black, PointSize[Large],
   Point[{{Cos[Pi/4], Sin[Pi/4], 0}, {Cos[Pi/4], Sin[Pi/4], 
      1}, {Cos[Pi/4], Sin[Pi/4], 2}}],
   myText["z = r", {Cos[35 °], Sin[35 °], 0.8}],
   myText["z = 2", {Cos[35 °], Sin[35 °], 1.8}]

   }], AxesLabel -> {"x", "y", "z"}, 
 PlotRange -> {{-2.5, 2.5}, {-2.5, 2.5}, {0, 2.5}}
 ]

For some reason, I am getting frequent kernel beeps when running the cell containing the image code. I don't know why.


enter image description here



Answer



There are two alternatives I would suggest, depending on what your plans for the Background are. Here is an illustration:


Text[Pane[Style["(1, 0, 0)", 12, Background -> Yellow], 
  ImageMargins -> 10], {1, 1, 0}]


small background


Text[Framed[Style["(1, 0, 0)", 12], Background -> Yellow, 
  FrameMargins -> 10, FrameStyle -> None], {1, 1, 0}]

big background


The first solution uses a Pane, but this doesn't have a Background option so that you have to specify the background color in the Style command, as you already did. I changed it to yellow just for clarity.


The second solution assumes that you want the background to fill out the extra space that was created, too. This can be done with Frame because it allows the Background option. I then set FrameStyle -> None to make the result look otherwise indistinguishable from a Pane.


When used in Graphics3D, these two outputs look as follows:


pane



enter image description here


To make this easier to use, just define something like this:


Options[myText] = {Background -> White, Padding -> 5, 
   FrameStyle -> None};

myText[s_, pos_, OptionsPattern[]] := 
 Text[Framed[Style[s, 12], Background -> OptionValue[Background], 
   FrameMargins -> OptionValue[Padding], 
   FrameStyle -> OptionValue[FrameStyle]], pos]


Graphics3D[myText["{1,0,0}", {0, 0, 0}], Background -> LightBlue]

function


The relevant options can be changed by adding them to in the invocation of myText. I used the name Padding for the option that you wanted, even though it's realized as FrameMargins internally.


Comments

Post a Comment

Popular posts from this blog

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

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

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