2
\$\begingroup\$

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.

Here 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? 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))
\$\endgroup\$
0

1 Answer 1

5
\$\begingroup\$

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)".

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.