Reduce works fine but it is more like fold-left. Is there any other form of reduce that lets me fold to right ?
Let's look at a possible definition of each:
(defn foldl [f val coll]
(if (empty? coll) val
(foldl f (f val (first coll)) (rest coll))))
(defn foldr [f val coll]
(if (empty? coll) val
(f (foldr f val (rest coll)) (first coll))))
Notice that only foldl
is in tail position, and the recursive call can be replaced by recur
. So with recur
, foldl
will not take up stack space, while foldr
will. That's why reduce
is like foldl
. Now let's try them out:
(foldl + 0 [1 2 3]) ;6
(foldl - 0 [1 2 3]) ;-6
(foldl conj [] [1 2 3]) ;[1 2 3]
(foldl conj '() [1 2 3]) ;(3 2 1)
(foldr + 0 [1 2 3]) ;6
(foldr - 0 [1 2 3]) ;-6
(foldr conj [] [1 2 3]) ;[3 2 1]
(foldr conj '() [1 2 3]) ;(1 2 3)
Is there some reason you want to fold right? I think the most common usage of foldr
is to put together a list from front to back. In Clojure we don't need that because we can just use a vector instead. Another choice to avoid stack overflow is to use a lazy sequence:
(defn make-list [coll]
(lazy-seq
(cons (first coll) (rest coll))))
So, if you want to fold right, some efficient alternatives are
reduced
to short-circuit reduce
.