How do I add a number, say 2, to every item in a list?
(setq x '(1 2))
(+ 2 x)
(mapcar '2+ x)
(loop for i in x
do (+ 2 i))
(dolist (i x)
(+ 2 i))
None of the above work.
Possible solutions:
(mapcar (lambda (entry) (+ entry 2)) x)
(mapcar (apply-partially #'+ 2) x)
And if you need to update x, then setq it to the result of one of the above forms, e.g.:
(setq x (mapcar (apply-partially #'+ 2) x))
That's not very elegant (there's probably a much simpler answer) but this should work:
(setq x '(1 2 3 4))
(defun add-one (x)
"Add 1 to each element of list X"
(setq counter 0)
(setq res x)
(while (< counter (length x))
(setcar (nthcdr counter res) (1+ (car (nthcdr counter x))))
(setq counter (1+ counter)))
res)
(add-one x)
-
1Never destructively modify (e.g. using
setcar) a quoted constant list, such as'(1 2 3 4), otherwise the constant will be modified globally and the same code may behave differently in each run. Instead, either initialisexto a new list, such as(list 1 2 3 4), before modifying it, or accumulate changes in a new list and set that as the value ofx, thus replacing its old value. – Basil Feb 11 '20 at 10:34 -
1Your looping logic is unidiomatic Elisp and inefficient, as it traverses the same list from the start for each element access. Idiomatic Elisp uses
dolist,mapc,mapcar, or successive calls tocdrto go from one list element to another, without the need for indices or traversing the entire length each time. Lists are not random-access like arrays. – Basil Feb 11 '20 at 10:38 -
(+ 2 x)doesn't work because+takes numbers as arguments and returns their sum, and(mapcar '2+ x)doesn't work because there is no built-in function called2+(though you could define it yourself). The two loops work, but you are not saving the results of their computation anywhere, so the results are lost. – Basil Feb 11 '20 at 10:24