Understanding mapcar

[CesarA]

I've seen mapcar in a few tutorials around. The way I see it, it applies a change to every element of a list.

 

For example (mapcar '1+ '(1 2 3))

would return something like (2 3 4)

 

And

(mapcar 'strcase '(a b c)) -> (A B C)

 

I understand this perfectly. But sometimes people use functions that are not that clear, for example

'cdr

 

I really don't get this, because cdr is supposed to return the 2nd element of a list. It's not a action I apply to a list, it's a return one. So how does (mapcar 'cdr ) works, and what does it do?

[rlx]

Hi CesarA,

 

 

You can also use foreach item (in) list do this or do that. I like to think of cdr as second and remaining. Mapcar 'cdr just means , for each element in list , or , apply the function cdr. If your list simply contains single elements this wouldn't work but if your list is like ((1 "a")(2 "b")...) then this function would return a list containing (("a") ("b") ...)

 

 

And if you use mapcar 'cadr instead of cdr you would get ("a" "b" ...)

 

 

gr.Rlx

Edited January 22, 2016 by rlx

[CesarA]

Sorry didn't quite understand this:

 

mapcar 'cadr returns "a" "b"?

 

but cadr is the 3rd element, and there's no 3rd element in ((1 "a")(2 "b")...)

 

how come 'cdr would return (("a") ("b") ...) (this i understand)

but 'cadr would return ("a" "b" ...)??

 

There's a difference, but, i don't think it makes sense

[rlx]

Sorry didn't quite understand this:

 

mapcar 'cadr returns "a" "b"?

 

but cadr is the 3rd element, and there's no 3rd element in ((1 "a")(2 "b")...)

 

how come 'cdr would return (("a") ("b") ...) (this i understand)

but 'cadr would return ("a" "b" ...)??

 

There's a difference, but, i don't think it makes sense

 

 

well , try it for yourself :

 

 

(setq lst '((1 "a")(2 "b")(3 "c")))
(mapcar 'cdr lst)
(mapcar 'cadr lst)

 

 

cdr will return a list with the first element removed (second and remaining)

cadr will give the first element of the second part , equal to (car (cdr ...)

 

 

There is a big difference between a list and a 'atom' , (list 1 2 3 4) is not the same as (list (list 1 2) (list 3 4))

 

 

the first is a list of 4 atoms , the second gives you a list of 2 list's , each list containing 2 atoms

 

 

gr. Rlx

[rlx]

try this

 

 

(defun c:tst ( / lst1 lst2 result1 result2 result3)
 ;age of jack and jill and their dog Bruno
 (setq lst1 '(("Jack" 22)("Jill" 25)("Bruno" 5))); -> (("Jack" 22) ("Jill" 25) ("bruno" 5))
 (setq result1 (car lst1)) ; -> ("Jack" 22)
 (setq result2 (cdr lst1)) ; -> (("Jill" 25) ("bruno" 5))
 (setq result3 (cadr lst1)); -> ("Jill" 25)
 (setq lst2 (list 1 2 3 4 5)); -> ( 1 2 3 4 5)
 (setq result1 (car lst2)) ; -> 1
 (setq result2 (cdr lst2)) ; -> (2 3 4 5)
 (setq result3 (cadr lst2)); -> 2 , same as (car (cdr lst2))
)

 

 

gr. Rlx

[CesarA]

I will do, thank you Rlx

[satishrajdev]

Go through this links :-

 

http://www.cadtutor.net/forum/showthread.php?80672-lambda-function-mapcar-apply-foreach-explained

 

http://www.cadtutor.net/forum/showthread.php?47026-Difference-between-foreach-and-mapcar