Skip to main content

list manipulation - Turning table into association for Classify


I have the following sample data for 20 users. I know their sex, and the response for three yes/no questions. I am trying to building a classifier to guess their response to the third question, based on their sex, and the response to the first two questions. It is probably obvious but this is all made up data.


Here is the data:


{{{"M", "Y", "N"} -> "Y"}, {{"F", "Y", "N"} -> 
   "N"}, {{"F", "N", "Y"} -> "Y"}, {{"M", "Y", "Y"} -> 
   "Y"}, {{"M", "Y", "Y"} -> "Y"}, {{"M", "N", "Y"} -> 

   "Y"}, {{"M", "Y", "Y"} -> "Y"}, {{"M", "N", "Y"} -> 
   "Y"}, {{"M", "Y", "N"} -> "Y"}, {{"F", "Y", "Y"} -> 
   "N"}, {{"F", "Y", "Y"} -> "N"}, {{"M", "Y", "N"} -> 
   "Y"}, {{"F", "Y", "N"} -> "N"}, {{"F", "Y", "N"} -> 
   "N"}, {{"M", "Y", "N"} -> "Y"}, {{"M", "Y", "N"} -> 
   "Y"}, {{"M", "Y", "Y"} -> "Y"}, {{"F", "Y", "Y"} -> 
   "N"}, {{"F", "N", "Y"} -> "N"}}

Then I make a training set from the first 10 elements of the data.


training = data[[1 ;; 10]]


and finally I try to use Classify using built in defaults


cf = Classify[training]

Which gives:


Argument {{{M,Y,N}->Y},{{F,Y,N}->N},{{F,N,Y}->Y},<<4>>,{{M,N,Y}->Y},{{M,Y,N}->Y},{{F,Y,Y}->N}} should be a rule, a list of rules, or an association. >>

I went through Iris data as practice, but since they uses built-in dataset, I'm not sure how the Associations were build behind the scenes.


I would like to understand:


1) How to convert my table, into associations? \ 2) In general, for Classifying is it better to use Mathematica's new Dataset structure?



This question does seems related to : Transform Dataset so it can be used as training set for Classify



Answer



You must change your list structure. right now it is in the form of a list of lists of associations. If you remove the lists, you will have a list of associations, which the classifier function can work with.


cf = Classify[Flatten[training,1]]

I would look at the documentation for AssociationMap and AssociationThread, for good examples of how to build associations.


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