Get first n elements of a list

Question

How can I get the first N elements of a list?

I know I can use subseq:

(subseq '("one" "two" "three" "four" "five") 0 2)

Yet when I use subseq, it produces one of the following results:

• on the first call, the message "function 'subseq' from cl package called at runtime" or,
• an error that "progn: Symbol’s function definition is void: subseq" when using vanilla emacs -Q.

I could overcome these by using (require 'cl-extras) or cl-subseq. But given these hiccups, I expect there is a more direct way to get the first N elements.

Looking through documentation didn't turn up anything and the implementation of cl-subseq is non-trivial. If a builtin doesn't exist, I imagine to implement from scratch would require something like "Get the car of each cdr". Yet, this makes me think of loop... which is also a cl extension.

Thoughts?

Answer 1

Use butlast. The following code snippet returns the first 3 elements, you can vary the number 3 in the snippet.

(let ((list '("one" "two" "three" "four" "five")))
  (butlast list (- (length list) 3)))

Answer 2

I think the sensible options are:

• (require 'cl-lib) and use cl-subseq
• (require 'seq) and use seq-subseq

The latter actually uses the former, but you might prefer using the seq library functions generally.

You could roll your own, but I would just use one of the above.

Answer 3

(Elevating ideasman42 comment to an answer)

Worth mention seq-take does more directly what the questioner asks.

(setq l1 '("one" "two" "three" "four" "five"))
(seq-take l1 3)
("one" "two" "three")