Clojure: finding sequential items from a sequence
Question
In a Clojure program, I have a sequence of numbers:
(2 3 4 6 8 1)
I want to find the longest sub-sequence where the items are sequential:
(2 3 4)
I am assuming that it will involve (take-while ...)
or (reduce ...)
.
Any ideas?
Clarification: I need the longest initial list of sequential items. Much easier, I'm sure. Thanks for the solutions to the more difficult problem I initially posed.
Solution
If you are only interested in the longest initial sequence, it's a 1-liner:
(defn longest-initial-sequence [[x :as s]]
(take-while identity (map #(#{%1} %2) s (iterate inc x))))
OTHER TIPS
Taking into account the OP's comment on the question -- which completely changes the game! -- this can be written very simply:
(let [doubletons (partition 2 1 [1 2 3 5 6])
increment? (fn increment? [[x y]]
(== (inc x) y))]
(cons (ffirst doubletons)
(map second (take-while increment? doubletons))))
;; returns (1 2 3)
Note that this is actually lazy. I expect it not to hold onto the head of doubletons
thanks to locals clearing. Another version:
(cons (first [1 2 3 5 6])
(map second (take-while increment? (partition 2 1 [1 2 3 5 6]))))
The original version of the question is more fun, though! :-) A super-simple solution to that could be built using the above, but of course that would be significantly less performant than using reduce
. I'll see if I have anything substantially different from zmila's and dnolen's solutions -- and yet still reasonably performant -- to add to that part of this thread later. (Not very likely, I guess.)
Answer to original:
(defn conj-if-sequential
([] [])
([a] a)
([a b] (let [a (if (vector? a) a [a])]
(if (= (inc (last a)) b)
(conj a b)
a))))
(reduce conj-if-sequential [2 3 4 6 8 1])
A more generic solution for those interested:
(defn sequential-seqs
([] [])
([a] a)
([a b] (let [n (last (last a))]
(if (and n (= (inc n) b))
(update-in a [(dec (count a))] conj b)
(conj a [b])))))
(defn largest
([] nil)
([a] a)
([a b] (if (> (count b) (count a)) b a)))
(reduce largest (reduce sequential-seqs [] [2 3 4 6 8 1 4 5 6 7 8 9 13]))
I think this is much better.
(defn find-max-seq [lst]
(let [[f & r] lst,
longest-seq (fn [a b] (if (> (count a) (count b)) a b)),
[last-seq max-seq] (reduce
(fn [ [[prev-num & _ :as cur-seq] max-seq] cur-num ]
(if (== (inc prev-num) cur-num)
[(conj cur-seq cur-num) max-seq]
[(list cur-num) (longest-seq cur-seq max-seq)]
))
[(list f) ()]
r)]
(reverse (longest-seq last-seq max-seq))))
(find-max-seq '(2 3 4 6 8 1)) ; ==> (2 3 4)
(find-max-seq '(3 2 3 4 6 8 9 10 11)) ; ==> (8 9 10 11)