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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.

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  • $\begingroup$ Consider that removing punctuation will make things like "it's" and "its" the same, although they obviously do not serve the same purpose. Is that intended / acceptable? $\endgroup$ – MarcoB Apr 29 '18 at 17:52
4
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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

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I've implemented the Bigrams and TF-IDF calculation according to @Henrik Schumacher suggestion (see the attached code). Thanks a lot, @Henrik Schumacher.

 bigramUniqe = <||>

 getbigrams[text_String] := Module[{words, res, temp},
 words = StringSplit[StringDelete[text, PunctuationCharacter]];
 temp = Partition[words, 2, 1];
 Scan[(If[MissingQ[bigramUniqe[#]], AssociateTo[bigramUniqe, # -> 1],
   AssociateTo[bigramUniqe, # -> (bigramUniqe[#] + 1)]]) &, temp];
 res = Counts[temp]
 ]

bigram=TF

 bigram = Table[getbigrams[data[[i]]], {i, 1, Length@data}]; 

bigramUniqe=iTF

 KeyValueMap[# ->  Log[2, Length@data/bigramUniqe[[Key[#]]]] &, bigramUniqe]
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  • 2
    $\begingroup$ If Henrik’s answer solved your problem the thing to do is accept it to show that this problem is solved and also give him a bit of a reward for being helpful. $\endgroup$ – b3m2a1 Apr 30 '18 at 16:19

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