问题
I'm trying to write a function with recur that cut the sequence as soon as it encounters a repetition ([1 2 3 1 4]
should return [1 2 3]
), this is my function:
(defn cut-at-repetition [a-seq]
(loop[[head & tail] a-seq, coll '()]
(if (empty? head)
coll
(if (contains? coll head)
coll
(recur (rest tail) (conj coll head))))))
The first problem is with the contains?
that throws an exception, I tried replacing it with some
but with no success. The second problem is in the recur
part which will also throw an exception
回答1:
You've made several mistakes:
- You've used
contains?
on a sequence. It only works on associative collections. Usesome
instead. - You've tested the first element of the sequence (
head
) forempty?
. Test the whole sequence. - Use a vector to accumulate the answer.
conj
adds elements to the front of a list, reversing the answer.
Correcting these, we get
(defn cut-at-repetition [a-seq]
(loop [[head & tail :as all] a-seq, coll []]
(if (empty? all)
coll
(if (some #(= head %) coll)
coll
(recur tail (conj coll head))))))
(cut-at-repetition [1 2 3 1 4])
=> [1 2 3]
The above works, but it's slow, since it scans the whole sequence for every absent element. So better use a set.
Let's call the function take-distinct
, since it is similar to take-while
. If we follow that precedent and make it lazy, we can do it thus:
(defn take-distinct [coll]
(letfn [(td [seen unseen]
(lazy-seq
(when-let [[x & xs] (seq unseen)]
(when-not (contains? seen x)
(cons x (td (conj seen x) xs))))))]
(td #{} coll)))
We get the expected results for finite sequences:
(map (juxt identity take-distinct) [[] (range 5) [2 3 2]]
=> ([[] nil] [(0 1 2 3 4) (0 1 2 3 4)] [[2 3 2] (2 3)])
And we can take as much as we need from an endless result:
(take 10 (take-distinct (range)))
=> (0 1 2 3 4 5 6 7 8 9)
I would call your eager version take-distinctv
, on the map
-> mapv
precedent. And I'd do it this way:
(defn take-distinctv [coll]
(loop [seen-vec [], seen-set #{}, unseen coll]
(if-let [[x & xs] (seq unseen)]
(if (contains? seen-set x)
seen-vec
(recur (conj seen-vec x) (conj seen-set x) xs))
seen-vec)))
Notice that we carry the seen elements twice:
- as a vector, to return as the solution; and
- as a set, to test for membership of.
Two of the three mistakes were commented on by @cfrick.
回答2:
There is a tradeoff between saving a line or two and making the logic as simple & explicit as possible. To make it as obvious as possible, I would do it something like this:
(defn cut-at-repetition
[values]
(loop [remaining-values values
result []]
(if (empty? remaining-values)
result
(let [found-values (into #{} result)
new-value (first remaining-values)]
(if (contains? found-values new-value)
result
(recur
(rest remaining-values)
(conj result new-value)))))))
(cut-at-repetition [1 2 3 1 4]) => [1 2 3]
Also, be sure to bookmark The Clojure Cheatsheet and always keep a browser tab open to it.
回答3:
I'd like to hear feedback on this utility function which I wrote for myself (uses filter
with stateful pred
instead of a loop
):
(defn my-distinct
"Returns distinct values from a seq, as defined by id-getter."
[id-getter coll]
(let [seen-ids (volatile! #{})
seen? (fn [id] (if-not (contains? @seen-ids id)
(vswap! seen-ids conj id)))]
(filter (comp seen? id-getter) coll)))
(my-distinct identity "abracadabra")
; (\a \b \r \c \d)
(->> (for [i (range 50)] {:id (mod (* i i) 21) :value i})
(my-distinct :id)
pprint)
; ({:id 0, :value 0}
; {:id 1, :value 1}
; {:id 4, :value 2}
; {:id 9, :value 3}
; {:id 16, :value 4}
; {:id 15, :value 6}
; {:id 7, :value 7}
; {:id 18, :value 9})
Docs of filter says "pred must be free of side-effects" but I'm not sure if it is ok in this case. Is filter
guaranteed to iterate over the sequence in order and not for example take skips forward?
来源:https://stackoverflow.com/questions/43839639/clojure-contains-conj-and-recur