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Difference between two maps

I need to very efficiently compare two maps in Clojure/Java, and return the difference as determined by Java's .equals(..), with nil/null equivalent to "not present".

i.e. I am looking for the most efficient way to a write a function like:

(map-difference
  {:a 1, :b nil, :c 2, :d 3}
  {:a 1, :b "Hidden", :c 3, :e 5})

=> {:b nil, :c 2, :d 3, :e nil}

I'd prefer an immutable Clojure map as output, but a Java map would also be fine if the performance improvement would be significant.

For what it's worth, my basic test case / expectation of behaviour is that the following will be equal (up to the equivalence of null = "Not present") for any two maps a and b:

a 
(merge b (difference a b))

What would be the best way to implement this?

like image 527
mikera Avatar asked Aug 02 '10 11:08

mikera


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1 Answers

I'm not sure what the absolutely most efficient way to do this is, but here's a couple of things which may be useful:

  1. The basic expectation of behaviour from the question text is impossible: if a and b are maps such that b contains at least one key not present in a, (merge b <sth>) cannot be equal to a.

  2. If you end up going with an interop solution but then need to go back to a PersistentHashMap at some point, there's always

    (clojure.lang.PersistentHashMap/create
     (doto (java.util.HashMap.)
       (.put :foo 1)
       (.put :bar 2)))
    ; => {:foo 1 :bar 2}
    
  3. If you need to pass the keyset of a Clojure map to a Java method, you can use

    (.keySet {:foo 1 :bar 2})
    ; => #< [:foo, :bar]>
    
  4. If all keys involved are guaranteed to be Comparable, this could be exploited for efficient computation of difference on maps with many keys (sort & merge scan). For unconstrained keys this is of course a no-go and for small maps it could actually hurt performance.

  5. It's good to have a version written in Clojure, if only to set a baseline performance expectation. Here is one: (updated)

    (defn map-difference [m1 m2]
            (loop [m (transient {})
                   ks (concat (keys m1) (keys m2))]
              (if-let [k (first ks)]
                (let [e1 (find m1 k)
                      e2 (find m2 k)]
                  (cond (and e1 e2 (not= (e1 1) (e2 1))) (recur (assoc! m k (e1 1)) (next ks))
                        (not e1) (recur (assoc! m k (e2 1)) (next ks))
                        (not e2) (recur (assoc! m k (e1 1)) (next ks))
                        :else    (recur m (next ks))))
                (persistent! m))))
    

    I think that just doing (concat (keys m1) (keys m2)) and possibly duplicating some work is likely more efficient most of the time than checking a given key is in "the other map" too at every step.

To wrap up the answer, here's a very simple-minded set-based version with the nice property that it says what it does -- if I misunderstood the spec, it should be readily apparent here. :-)

(defn map-difference [m1 m2]
  (let [ks1 (set (keys m1))
        ks2 (set (keys m2))
        ks1-ks2 (set/difference ks1 ks2)
        ks2-ks1 (set/difference ks2 ks1)
        ks1*ks2 (set/intersection ks1 ks2)]
    (merge (select-keys m1 ks1-ks2)
           (select-keys m2 ks2-ks1)
           (select-keys m1
                        (remove (fn [k] (= (m1 k) (m2 k)))
                                ks1*ks2)))))
like image 67
Michał Marczyk Avatar answered Sep 30 '22 15:09

Michał Marczyk