Skip to main content

plotting - Getting over Mathematica Bar Chart Limitations….


I am trying to create a pareto chart for some non technical folks and having the hardest time formatting it correctly…


Here’s my sample code:


(*Some dynamically generated data *)

mydata = {0.9436, 2.20725333345, 2.1402, 1.8729, 4.9436, 0.4819};
mylabels = {"l1", "l2", "l3", "l4", "some really long label", "another really long label", "l7"};


(*lets riffle and sort the data *)

a =  Sort[ Partition[Riffle[mylabels, mydata], 2], #1[[2]] > #2[[2]] & ];

BarChart[a , BarOrigin -> Left, BarSpacing -> -1, LabelingFunction -> (Placed[#, After] &), Axes -> None, Frame -> True, FrameTicks -> None, FrameLabel -> "Some Title", GridLines -> None, AspectRatio -> 0.3, ImageSize -> Full, ChartStyle -> "Pastel"]

A few questions:





  • What is the best way to truncate to 2 digits in the labels, I tried using NumberForm in the Sort function but doesn’t work since Numberform is a wrapper.




  • Labeling works reasonably well when labels are short but I seem to run into placement issues with long labels… Any ideas on better ways to handle this?




  • Frame produced by Mathematica seems to overlap with the labels…




Any suggestions gladly appreciated!




Answer



Here's one approach.


mydata = N[Round[{0.9436, 2.20725333345, 2.1402, 1.8729, 4.9436, 0.4819}*100]/100];

This will make the data appear with only 2 decimal places of accuracy. If you are willing to rearrange things a bit, it's pretty easy to remove the overlap...


BarChart[Labeled[#2, #1, Before] & @@@ a, BarSpacing -> 0, 
  LabelingFunction -> (Placed[#, After] &), BarOrigin -> Left, 
  PlotLabel -> "Some Title"]

This gives:



enter image description here


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