Skip to main content

list manipulation - Bigrams and TF-IDF calculation



I want to create a bag of bigrams in a set of documents and calculate the TF-IDF vector of each document. To calculate the bigram of the text I used the following code: The small example of the data (each element in the list is a different document)


data = {"The food at snack is a selection of popular Greek dishes. 
The appetizer tray is good as is the Greek salad. We were underwhelmed with 
the main courses. There are 4-5 tables here so it's 
sometimes hard to get seated.","This little place in Soho is wonderful. I 
had 
a lamb sandwich and a glass of wine. The price shocked me for how small the 
serving was, but then again, this is Soho. The staff can be a little snotty 
and rude, but the food is great, just don't expect world-class service.", 
"ordered lunch for 15 from Snack last Friday.  On time, nothing 

missing and the food was great.  I have added it to the regular 
company lunch list, as everyone enjoyed their meal."}

The way that I create the bigrams in the file (set of documents) and calculate the TF for each bigram in specific document:


bigram =
  Table[
    Merge[
      <|First[#] <> " " <> Last[#] -> 1|> & /@ 
        Partition[
          StringSplit[StringReplace[data[[i]], PunctuationCharacter ->""]],

          2, 1],
      Total],
    {i, 1, Length@data}]

The way that I calculate the frequency of the bigrams in the file (ITF):


bigramUniqe = <||>
Scan[(If[MissingQ[bigramUniqe[#]], AssociateTo[bigramUniqe, # -> 1], 
 AssociateTo[bigramUniqe, # -> (bigramUniqe[#] + 1)]]) &, bigram];

But in this way, I do not succeed to count the frequency of document that contains the specific bigrams( I have some issue with the level specification of the Associate). Anyway, I look for a more efficient way to implement this task.Thank in advance for any suggestions.




Answer



What do you think about skipping the StringJoin and storing a bigram as a pair of strings?


getbigrams[text_String] := Module[{words},
  words = 
   StringSplit[
    ToLowerCase[StringDelete[text, PunctuationCharacter]]];
  Counts[Partition[words, 2, 1]]
  ]

That can save about 40 % of time:



data = ExampleData /@ ExampleData["Text"];

a = Table[
     Merge[<|First[#] <> " " <> Last[#] -> 1|> & /@ 
       Partition[
        StringSplit[
         StringReplace[ToLowerCase[data[[i]]], 
          PunctuationCharacter -> ""]], 2, 1], Total], {i, 1, 
      Length@data}]; // AbsoluteTiming // First


b = getbigrams /@ data; // AbsoluteTiming // First

Values[a] == Values[b]


7.62668


4.40748


True



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

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

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