Given a sequence of items I want to find the n most frequent items, in descending order of frequency. So for example I would like this unit test to pass:
(fact "can find 2 most common items in a sequence"
(most-frequent-n 2 ["a" "bb" "a" "x" "bb" "ccc" "dddd" "dddd" "bb" "dddd" "bb"])
=>
'("bb" "dddd"))
I am fairly new to Clojure and still trying to get to grip with the standard library. Here is what I came up with:
(defn- sort-by-val [s] (sort-by val s))
(defn- first-elements [pairs] (map #(get % 0) pairs))
(defn most-frequent-n [n items]
"return the most common n items, e.g.
(most-frequent-n 2 [:a :b :a :d :x :b :c :d :d :b :d :b]) =>
=> (:d :b)"
(take n (->
items ; [:a :b :a :d :x :b :c :d :d :b :d :b]
frequencies ; {:a 2, :b 4, :d 4, :x 1, :c 1}
seq ; ([:a 2] [:b 4] [:d 4] [:x 1] [:c 1])
sort-by-val ; ([:x 1] [:c 1] [:a 2] [:b 4] [:d 4])
reverse ; ([:d 4] [:b 4] [:a 2] [:c 1] [:x 1])
first-elements))) ; (:d :b :a :c :x)
However this seems like a complicated chain of functions to do a fairly common operation. Is there a more elegant or more idiomatic (or more efficient) way to do this?
As you have discovered, typically you would use a combination of sort-by and frequencies to get a frequency-sorted list.
(sort-by val (frequencies ["a" "bb" "a" "x" "bb" "ccc" "dddd" "dddd" "bb" "dddd" "bb"]))
=> (["x" 1] ["ccc" 1] ["a" 2] ["dddd" 3] ["bb" 4])
Then you can manipulate this fairly easily to get the lowest / highest frequency items. Perhaps something like:
(defn most-frequent-n [n items]
(->> items
frequencies
(sort-by val)
reverse
(take n)
(map first)))
Which again is pretty similar to your solution (apart from that you don't need the helper functions with clever use of the ->>
macro).
So overall I think your solution is pretty good. Don't worry about the chain of functions - it's actually a very short solution for what is logically quite a complicated concept. Try coding the same thing in C# / Java and you will see what I mean......
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