The question about "apply" , "mapcar" .

Hello everyone.

I have a questionabout "apply" and "mapcar".

Please looks bellow code:

(setq b '((0 1 2) (3 4 5) (6 7 ))
(apply 'mapcar (cons 'list b))

it return : ((0 3 6 ) (1 4 7 ) (2 5 8 ) )

how it works?

and next question:

(setq a '(1 2 3 4 5 6 7 8 9 0))
(apply 'mapcar (cons 'list (list a (cdr a))))

it return : ( (1 2) (2 3) (3 4) (4 5) (5 6) (6 7) (7 8 ) (8 9) (9 0) )

how it works also?

Welcome to CADTutor .

Hope that I could describe them as clear as you looking for

The first function apply implement the mapcar function on a list of lists and mapcar function iterate through each element of every list and with the use of cons and list functions you have them gathered in a list of lists as well .

The second question is also implement the mapcar function with the use of apply function to get the first and the second item of the list separately and I can offer you as alternative function in recursive manner.

(defun _list  (l)
;;;  Tharwat . 05.02.2015    ;;;
 (if (> (length l) 1)
   (cons (list (car l) (cadr l)) (_list (cdr l))))
 )

To test the above function :

(_list '(1 2 3 4 5 6 7 8 9 0))

[color=red];; Returns [/color]
((1 2) (2 3) (3 4) (4 5) (5 6) (6 7) (7  (8 9) (9 0))

Thanks to you, Mr.Tharwat !

But i still confuse on some details.

First,

I know the mapcar function just like foreach. In code 1 , why the next value is "3" not "1" ?

please looks other example :

(setq lst '((3 4 5) (0 1 2) (6 7 ))
(apply 'mapcar (cons 'min lst))

the result shows (0 1 2)

why its not (3 0 6)?

Second,

Why add the list(using apostrope for not to evulate) in fornt of a list , and using apply can gather the list?

Thanks!

Thanks to you, Mr.Tharwat !

But i still confuse on some details.

First,

I know the mapcar function just like foreach. In code 1 , why the next value is "3" not "1" ?

Because the mapcar function iterate through each first element of the target list first then the second then the third .... to the end .

(setq lst '((3 4 5) (0 1 2) (6 7 ))
(apply 'mapcar (cons 'min lst))

the result shows (0 1 2)

why its not (3 0 6)?

Because the minimum values for the target list is consist of three element and the outcome came in three including the minimum numbers .

Second,

Why add the list(using apostrope for not to evulate) in fornt of a list , and using apply can gather the list?

That required from the mapcar function , first apply function needs a function as a first argument then the mapcar function also requires a function as a first argument .

I am happy with your questions because that also refresh my memory with the fore-said functions

Thanks you !

I have clear understand those function after yours explaination!

Thanks you !

I have clear understand those function after yours explaination!

happy to hear that and you are most welcome .

Keep on practicing with functions to become familiar with them and don't be surprised or panic if you one day have noticed that a specific function doing something than you

used to see or to know an example one is the entdel function .

Please looks bellow code:

(setq b '((0 1 2) (3 4 5) (6 7 ))
(apply 'mapcar (cons 'list b))

it return : ((0 3 6 ) (1 4 7 ) (2 5 8 ) )

 

how it works?

Consider that:

(apply 'mapcar (cons 'list '((0 1 2) (3 4 5) (6 7 )))

Is the same as:

(mapcar 'list '(0 1 2) '(3 4 5) '(6 7 )

Which evaluates in the same way as:

(list (list 0 3 6) (list 1 4 7) (list 2 5 )

Returning:

((0 3 6) (1 4 7) (2 5 )

Thanks you also , Mr. Lee Mac!

I try other example, it really amazing.

But why using apply and mapcar , why not just using mapcar?

Is that any reasons ? much faster ?

Thanks!

Thanks you also , Mr. Lee Mac!

You're welcome!

But why using apply and mapcar , why not just using mapcar?

Is that any reasons ? much faster ?

The function or functions required depends on the format of the input data, the operations that are to be performed, and the required format for the list that is returned - in your particular example, if each list of three elements is nested within a parent list, this list would be supplied to mapcar as a single argument and would not return the same result:

_$ (mapcar 'list '((0 1 2) (3 4 5) (6 7 ))
(((0 1 2)) ((3 4 5)) ((6 7 ))

Hence the apply function is used to construct an expression equivalent to:

(mapcar 'list '(0 1 2) '(3 4 5) '(6 7 )