Clojure basics: counting frequencies

Question

I am learning Clojure, and I saw this bit of code online:

(count (filter #{42} coll))

And it does, as stated, count occurrences of the number 42 in coll. Is #{42} a function? The Clojure documentation on filter says that it should be, since the snippet works as advertised. I just have no idea how it works. If someone could clarify this for me, that would be great. My own solution to this same thing would have been:

(count (filter #(= %1 42) coll))

How come my filtering function has parenthesis and the snippet I found online has curly braces around the filtering function (#(...) vs. #{...})?

Have you seen this answer? http://stackoverflow.com/questions/3249334/test-whether-a-list-contains-a-specific-value-in-clojure — Jonas, May 03 '14 at 18:13
No, I have not seen that answer. I can see that they are also using similar bits of code. There is also mention that `(#{x} x)` returns `x`. I just don't get what `#{x}` is doing as a function. Isn't that just a set? — vim, May 03 '14 at 18:24
See my answer. A set is not just a set - amongst other things it's an IFn so you can treat it as a function. — pete23, May 03 '14 at 18:25
This is tangentially related, but in case you weren't aware, Clojure also has a function called [frequencies](http://clojuredocs.org/clojure_core/clojure.core/frequencies) that is useful for cases like this. It produces a map of each item in a collection to the number of times it occurs. So you could do the above like this: `((frequencies coll) 42)` -- `(frequencies coll)` produces a map, which is used as a function to look up `42` and return the number of occurrences. (EDIT: To be fair, this is not as performant) — Dave Yarwood, May 05 '14 at 15:02

pete23 · Accepted Answer · 2014-05-04T06:49:55.543

=> #{42}
#{42}

Defines a set...

=> (type #{42})
clojure.lang.PersistentHashSet

=> (supers (type #{42}))
#{clojure.lang.IHashEq java.lang.Object clojure.lang.IFn ...}

Interestingly the set implements IFn so you can treat it like a function. The behaviour of the function is "if this item exists in the set, return it".

=> (#{2 3} 3)
3
=> (#{2 3} 4)
nil

Other collections such as map and vector stand in as functions in a similar fashion, retrieving by key or index as appropriate.

=> ({:x 23 :y 26} :y)
26
=> ([5 7 9] 1)
7

Sweet, no? :-)

score 1 · Answer 2 · answered May 07 '14 at 08:50

Yes, #{42} is a function,

because it's a set, and sets, amongst other capabilities, are functions: they implement the clojure.lang.IFn interface.
Applied to any value in the set, they return it; applied to anything else, they return nil.
So #{42} tests whether its argument is 42 (only nil and false are false, remember).

The Clojure way is to make everything a function that might usefully be one:

Sets work as a test for membership.
Maps work as key lookup.
Vectors work as index lookup.
Keywords work as lookup in the map argument.

This

often saves you a get,
allows you, as in the question, to pass naked data structures to higher order functions such as filter and map, and
in the case of keywords, allows you to move transparently between maps and records for holding your data.

Clojure basics: counting frequencies

2 Answers2