Monday, August 30, 2010

Clojure: Using Sets and Maps as Functions

Clojure sets and maps are functions.

Since they are functions, you don't need functions to get values out of them. You can use the map or set as the example below shows.
(#{1 2} 1)
> 1

({:a 2 :b 3} :a)
> 2
That's nice, but it's not exactly game changing. However, when you use sets or maps with high order functions you can get a lot of power with a little code.

For example, the following code removes all of the elements of a vector if the element is also in the set.
(def banned #{"Steve" "Michael"})
(def guest-list ["Brian" "Josh" "Steve"])

(remove banned guest-list)
> ("Brian" "Josh")
I'm a big fan of using sets in the way described above, but I don't often find myself using maps in the same way. The following code works, but I rarely use maps as predicates.
(def banned {"Steve" [] "Michael" []})
(def guest-list ["Brian" "Josh" "Steve"])

(remove banned guest-list)
> ("Brian" "Josh")
However, yesterday I needed to compare two maps and get the list of ids in the second map where the quantities didn't match the quantities in the first map. I started by using filter and defining a function that checks if the quantities are not equal. The following code shows solving the problem with that approach.
; key/value pairs representing order-id and order-quantity
(def map1 {1 44 2 33})
(def map2 {1 55 2 33})

(defn not=quantities [[id qty]] (not= (map1 id) qty))
(keys (filter not=quantities map2))
> (1)
However, since you can use maps as filter functions you can also solve the problem by merging the maps with not= and filtering by the result. The following code shows an example of merging and using the result as the predicate.
; key/value pairs representing order-id and order-quantity
(def map1 {1 44 2 33})
(def map2 {1 55 2 33})

(filter (merge-with not= map1 map2) (keys map2))
> (1)
I don't often find myself using maps as predicates, but in certain cases it's exactly what I need.

1 comment:

  1. merge-with is probably my favorite map function.

    ReplyDelete

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