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up vote 2 down vote favorite

I wanted to do the following:

  • Count the frequencies of words in a text (over 5 letters)
  • Invert the map of words to frequencies, but group together words that have the same frequency in the inversion.
  • Sort the inverted map by keys descending order and take the top 25.

Below is the code I came up with. Did I re-invent the wheel with map-invert-preserve-dups? Is there a more concise way to do anything I did below? Am I doing anything unnecessarily (i.e. `(~k)?

(defn map-invert-preserve-dups
  [m]
  (reduce
    (fn [m [k v]]
      (if (contains? m v)
        (assoc m v (cons k (get m v)))
        (assoc m v `(~k))))
    {}
    m))

(->> "http://www.weeklyscript.com/Pulp%20Fiction.txt"
  (slurp)
  (re-seq #"\w{5,}")
  (frequencies)
  (map-invert-preserve-dups)
  (sort)
  (reverse)
  (take 25))
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migrated from stackoverflow.com Dec 10 '11 at 14:12

This question came from our site for professional and enthusiast programmers.

1 Answer 1

up vote 5 down vote accepted

Well, the most obvious fix is indeed map-invert-preserving-dups - the whole thing could be more easily written as:

(defn map-invert-preserving-dups [m]
  (apply merge-with into
         (for [[k v] m]
           {v [k]})))

The for expression yields a sequence of maps like [{a 1} {b 2} {a 5}]. Apply calls merge-with into on all of those maps. If you look up the definition of merge-with, you can see that this means basically: "Merge all of these maps together, and if the same key exists twice, with values x and y, then make its value (into x y)".

share|improve this answer
    
Thanks sure your answer. Do mind adding a concise explanation of how the expression "apply merge-with into" fits together? –  noahz Dec 11 '11 at 19:33
1  
@noahz Edited question with explanation. –  amalloy Dec 12 '11 at 2:14
    
Awesome, thanks! –  noahz Dec 12 '11 at 18:52

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