Skip to main content

plotting - Styling ticks in a plot



I managed to turn the ticks inside-out but the ticks go now through the numbers. In fact I would like to have it like in the picture attached (ticks pointing outside the plot and no additional ticks between the numbers). Has anyone a suggestion?


data={-1.2056,-(7288960271654/4985881289361),-1.30053,-(33199059446612625/13128459048331036),-(5130109517723890585/5148852459134416857),-1.73904,-1.164,-1.83398,-0.97505,-(152151059187768664247/302333262014074402254),-(33596749386427335752225/52657796909983176600514),-0.785963,-0.398132,-(2414923435622638997136189344226575/3392598745831412249336953513454508),-0.820439};
a=First[HistogramList[data]]
b=Range[Max[Last[HistogramList[data]]]]
Histogram[
  data,
  BarOrigin->Left,
  Axes->False,
  Frame->{{True,None},{True,None}},
  FrameTicks->{

              {Transpose[{a,a,Table[{-0.02,0},{i,Length[a]}]}],None},
              {Transpose[{b,b,Table[{-0.02,0},{i,b}]}],None}
              }
]

Example



Answer



This works, but it does require some manual tweaking


Histogram[data, BarOrigin -> Left, Axes -> False, 
 Frame -> {{True, None}, {True, None}}, 

 FrameTicks -> {{Transpose[{a, 
      Row[{#, Pane["", {10, Automatic}]}] & /@ a, 
      Table[{-0.02, 0}, {i, Length[a]}]}], 
    None}, {Transpose[{b, 
      Pane[#, {Automatic, 20}, Alignment -> {Center, Bottom}] & /@ b, 
      Table[{-0.02, 0}, {i, b}]}], None}}]

Mathematica graphics


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