Counting occurences (solution to the exercise)

unsorted — cgrand, 28 April 2009 @ 18 h 00 min
What does (merge-with + {:a 12} {:b 4} {:a 3 :b 7}) return?
{:b 11, :a 15} when there are several values for a key, these values are merged (two at a time) using the specified function — here they are summed.
Can you count occurrences of each value in a collection using merge-with?
(apply merge-with + (map (fn [x] {x 1}) coll)) or, using for: (apply merge-with + (for [x coll] {x 1})).

Counting occurences

unsorted — cgrand, 27 April 2009 @ 17 h 44 min

One asked me how to count occurrences of each value in a collection. I answered (reduce #(assoc %1 %2 (inc (%1 %2 0))) {} coll). Since it can take some time to get accustomed to the functional way of thought, here is how one can work such an expression out:

How to count all occurrences of 42?
(count (filter #{42} coll))
How to express count using reduce?
(defn my-count [coll] (reduce (fn [n _] (inc n)) 0 coll))
How to count all occurrences of 42 using reduce?
(reduce (fn [n _] (inc n)) 0 (filter #{42} coll))
Can you get rid of the filter?
(reduce (fn [n x] (if (= 42 x) (inc n) n)) 0 coll)
I’d like the result to be {42 occurences-count}.
(reduce (fn [m x] (if (= 42 x) (assoc m 42 (inc (m 42))) m)) {42 0} coll)
42 is hardcoded in four places, it’s bad!
(reduce (fn [m x] (if (= 42 x) (assoc m x (inc (m x))) m)) {42 0} coll)
Can’t you replace {42 0} with {}?
No (inc (m x)) would fail because (m x) would return nil.
How does one provide a default value when the key is not found ?
(a-map a-key default-value)
Can’t you replace {42 0} with {}?
(reduce (fn [m x] (if (= 42 x) (assoc m x (inc (m x 0))) m)) {} coll)
I changed my mind I’d like you to count occurrences of each value.
Easy! (reduce (fn [m x] (assoc m x (inc (m x 0)))) {} coll) or, terser, (reduce #(assoc %1 %2 (inc (%1 %2 0))) {} coll)

Exercise:

What does (merge-with + {:a 12} {:b 4} {:a 3 :b 7}) return?
Can you count occurrences of each value in a collection using merge-with?
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