Skip to main content

graphics - How can I extract the maximum value from a Histogram?


In Mathematica 8.04 I've created a histogram of returns on a stock using:


returns = FinancialData["SP500", "Return", {Date[] - {5, 0, 0, 0, 0, 0}, Date[]}, "Value"];
μ = Mean[returns];
h = Histogram[returns, 300, "PDF"];

Mathematica graphics



where "returns" is a list of the daily returns, and I want 300 bins in the probability density method. Next, I would like to draw a line that represents the mean of the returns so I use:


meanLine = Graphics[{Thick, Darker[Green], Line[{{μ, 0}, {μ, maxFreq + 2}}]}];

The problem is that I need to calculate the highest y-value for the line, which should be equal to the count of the data points in the bin with the most points (I've called this maxFreq above). I am currently setting this value manually because I don't know how to extract it from the histogram (h). Note that the x-coordinate of this line (μ) is simply the mean of the returns.


I've tried looking at FullForm[h] to see if I could figure out how to extract the data. Buried within that output is the following:


List[List[Rectangle[List[-0.0015, 0.], List[-0.001, 54.0111]

which shows that I want to set maxFreq = 54.0111. So, I suppose that I want to use the Max function, but I don't know how to access the heights of each bin. I think that I need to use some variation of the Part function, but I can't figure out how. Any clues would be appreciated.


Here is the code that will generate what I'm looking for, but maxFreq is manually entered:


returns = FinancialData["SP500", "Return", {Date[] - {5, 0, 0, 0, 0, 0}, Date[]}, "Value"];

μ = Mean[returns];
h = Histogram[returns, 300, "PDF"];
maxFreq = 54;
meanLine = Graphics[{Thick, Darker[Green],Line[{{μ, 0}, {μ, maxFreq + 2}}]}];
Show[h, meanLine]

Answer



As mentioned by others, use HistogramList. You can even use the resulting information to generate the plot without recomputing the information:


{bins, heights} = HistogramList[returns, 300, "PDF"];
maxFreq = Max[heights];
Histogram[returns, {bins}, heights &, 

 Epilog -> {{Thick, Darker[Green], 
    Line[{{μ, 0}, {μ, maxFreq + 2}}]}}]

enter image description here


Comments

Post a Comment

Popular posts from this blog

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
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