Returning first k elements of a list. My code returns last k elements

Question

I am trying to return the first k elements of a list xs. For example take 2 [1; 2; 3] should output [1; 2].

I wrote a function, but instead of returning the first k elements it returns the last k elements, and instead of returning the whole list if k equals to length of the list, it fails.

This is my code

let rec take k = function
|[] -> failwith "take"
|x :: xs -> if k = List.length xs  then xs else take (List.length xs - k) xs

How can I fix this?

see https://stackoverflow.com/questions/19944969/prolog-accumulators-are-they-really-a-different-concept/19951540#19951540 for some (hopefully helpful) verbiage. It is language-agnostic, contrary to the title. (also, [this](https://stackoverflow.com/questions/27803152/sicp-example-counting-change-cannot-understand/27804489#27804489) or [this](https://stackoverflow.com/questions/12659581/functional-programming-lots-of-emphasis-on-recursion-why/12662393#12662393)). — Will Ness, Oct 21 '16 at 08:24

Jeffrey Scofield · Accepted Answer · 2016-10-21T16:12:03.737

You need to answer the two basic questions for writing recursively:

What cases are so trivial that the answer is obvious?
If given a non-trivial case, how do I make a smaller instance of it, so I can get the answer for that smaller instance just by using this same function that I am writing¹, and then find my full answer from that smaller answer by some simple steps?

For this problem, the trivial cases (I would say) are when k is 0, and when the given list is an empty list.

For the non trivial case, the insight seems to be that I can get the first k elements of a list (for k > 0) by getting the first (k - 1) elements of the tail of the list, and then adding the head of the original list onto that&hairsp;².

This is not what your code does, so that's basically why it doesn't work :-)

As a side comment, I wouldn't consider it an error to ask for the first 0 elements of any list (including the empty list).

¹ and the essence of recursion method is to assume you already have it at your disposal.

² this is because of the identity length (x :: xs) = 1 + length xs.

So I should replace failwith with if k = 0 then [] , and for the second line if k = 0 then [] else take (k-1) xs @ [x] ? I will give it a look again and try to solve it. Right now I can hardly understand any of this :-). Thank you. — Uysal M, Oct 20 '16 at 23:30
You want to put the head back on the beginning. `take (k - 1) xs @ [x]` puts it at the end. Hint: use the `::` operator. As a side comment, learning to think recursively is a real pleasure (IMHO). — Jeffrey Scofield, Oct 20 '16 at 23:32

score 0 · Answer 2 · answered Oct 21 '16 at 18:58

Just a different way :

let take k l =
  let v =Array.of_list l in
  let v'=Array.make k 0 in
  Array.blit v 0 v' 0 k;
  Array.to_list v';;

test:

# take 2 [1;2;3];;
- : int list = [1; 2]
# take 0 [1;2;3];;
- : int list = []
# take 4 [1;2;3];;
Exception: Invalid_argument "Array.blit".

Returning first k elements of a list. My code returns last k elements

2 Answers2