Clojure - Splitting a sequence into multiple sequences

2
votes

I am working on a function that will split a sequence of dates (or anything else) into a number of sequences contained within a vector based on a given number x.

   (date1 date2 date3 date4 date5 date6 date7)

So given the list of dates above and passing in the variable 2 it will produce the vector below.

   [(date1 date2) (date3 date4) (date5 date6) (date7)]

The code I have so far is below but all it returns is a vector containing nil.

(defn date-splitter [date-count dates x]
  (loop [i date-count, current-split dates, split-dates (vector)]
        (if (<= i x)
            (conj split-dates (get current-split 1))
            (let [s (split-at x current-split)]
              (recur (int (- i x)) (get s 1) (conj split-dates (get s 0)))))))

I have also had a look at the split-with function, thinking that I could use it to split the sequence when the modulus of the index divided by x is zero, but I haven't had any luck with this.

Any help would be greatly appreciated.

David.

2
clojure

2 Answers

4
votes

Take a look at split's cousin partition

=> (partition-all 2 '(1 2 3 4 5 6 7))
((1 2) (3 4) (5 6) (7))

If you want to do index based operations, you can use map-indexed and keep-indexed.

=> (map-indexed (fn [idx itm] [(Math/floor (/ idx 2)) itm]) [1 2 3 4 5 6 7])
([0.0 1] [0.0 2] [1.0 3] [1.0 4] [2.0 5] [2.0 6] [3.0 7])

In FP, non-index based operations are usually the better option though. Index based solutions can be considered code smell.

There's several ways of doing partition the FP way without index. Ankur's is a great example of how to solve this if partition-all wouldn't have been in core clojure.

2
votes
(defn date-splitter [n v]
  (if (not (seq v)) 
      [] 
      (lazy-cat [(take n v)] (date-splitter n (drop n v)))))


user=> (date-splitter 3 [1,2,3,4])
((1 2 3) (4))