how to covert list

What functions should i be looking for to convert a list in the format:

("A") ("B") ("C") ("D")

to the format:

("A" "B" "C" "D")

It's probably something simple but I'm still learning

Thanks!

 

Or (append '("A") '("B") '("C") '("D")).

You're not converting a list, you're merging 4 lists into one.

 

Or rather (apply 'append '(1 2 3 4 5))

 

Or rather (apply 'append '(1 2 3 4 5))

That isn't valid, and it doesn't address the original question, merging
multiple lists into one.

My suggestion (append '("A") '("B") '("C") '("D")) correctly evaluates to
("A" "B" "C" "D") which is what the poster wanted.

 

$ (apply 'append '(("A") ("B") ("C") ("D")))

That's correct if there is actually a list of lists to begin with -- the
original post wasn't clear. I assumed the goal was to merge separate lists.

 

Silly me,

Or rather (apply 'append '((a) (b) (c) (d)))

The poster's question was directed at a single list per the statement:

I guess I read between the lines, but how couldn't I when he said list and
wrote ("A") ("B") ("C") ("D")?

And yes I know your answers works and is good I was simply stating an
alternative that his statement could have been addressing.

 

Thanks everyone - You found more than one solution for me, now I get to play
with 'em and see which I like best.

kemp

 

Hi Luis and all,
Now who would like to show how to flatten any nested list
such as '((1 . 2) (1 2 3) ((1 2) (2 3) 4))

into '(1 2 1 2 3 1 2 2 3 4)

One issue would be should nil elements remain? I won't
take a position on that one.

Kudos for best iterative solution, best recursive solution,
fastest and wittiest. Links to previous discussions are
welcome.

Regards,
Doug

 

based on your example you could do this...

(setq lst '((1 . 2) (1 2 3) ((1 2) (2 3) 4)))

(defun en ( lst / pass )
(traverse_list lst)
pass
)

(defun traverse_list ( lsts / a p )
(mapcar
'(lambda (a)
(if
(listp a)
(traverse_list
(vl-remove-if 'null
(append
(list (car a))
(if (listp (setq p (cdr a))) p (list p))
)
)
)
(progn
(setq pass (append pass (list a)))
a
)
)
)
lsts
)
)

try (en lst)

HTH,

Devin

 

Looks good Devin. Lets see if anyone else has a different
approach.

<snip>

 

This is simply a quick rewrite of an elegant recursive
function posted by ____ _____ that used a cond structure;
this one just uses if statements and makes one improvement
on the original resulting in ~ 7% performance gain (very
casual / quick test).

(defun delist ( lst / x )
(if lst
(if (atom lst)
(list lst)
(if (atom (setq x (car lst)))
(cons x (delist (cdr lst)))
(append (delist x) (delist (cdr lst)))
)
)
)
)

What's the one improvement?

Does it still work?

(delist '((1 . 2) (1 2 3) ((1 2) (2 3) 4)))

(1 2 1 2 3 1 2 2 3 4)

Just havin' fun

PS: Gues who ____ _____ is?

Hi Luis and all,
Now who would like to show how to flatten any nested list
such as '((1 . 2) (1 2 3) ((1 2) (2 3) 4))

into '(1 2 1 2 3 1 2 2 3 4)

One issue would be should nil elements remain? I won't
take a position on that one.

Kudos for best iterative solution, best recursive solution,
fastest and wittiest. Links to previous discussions are
welcome.

Regards,
Doug

 

Hi Michael,

Doug, and it was called flatten.

You think we don't save these gems, Doug? ;-)

Jeo Burke

 

Have you changed your name? Struck me
as funny when I read it

 

Jason,

That's the name I use on my tax return. ;-)

 

Truly a gem.

 

<snicker>

 

Thanks Guys,

Must be early Alzheimer's. I had forgotten that I had
posted one already.

Michael. Thanks for posting your adaptation. I
must have called it something else on my system
cause my current flatten zaps the z coordinates on
everything to 0.0.

G'nite

 

(defun flatten (lst)
(cond ((null lst) nil)
((atom lst) (list lst))
(T (append (flatten (car lst))
(flatten (cdr lst))))))

- sorry about possible indentation oddities, having trouble with fonts.

--

 

The weird thing is mine is called delist for the very same reason!

<xfiles.wav>

Thanks Guys,

Must be early Alzheimer's. I had forgotten that I had
posted one already.

Michael. Thanks for posting your adaptation. I
must have called it something else on my system
cause my current flatten zaps the z coordinates on
everything to 0.0.

G'nite

 

.... mine is renamed "flatlist" for the same reason.

Martti's post is interesting. It looks like Doug's function, while leaving out his
third condition. And it seems to work OK.

Here's what I have from the original thread.

; D. C. Broad, Jr. 10/25/02 - was named "flatten" - renamed flatlist
; in list:
; '( 0 (1 2) ((3 . 4) (5 6)) (((7 8) (9 10 11 12)))
; ((((13 . 14) (15 16) (17 . 18) (19 20)))) (21)))
; out list:
; (0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21)
(defun flatlist (l)
(cond
((null l) nil)
((atom l) (list l))
((atom (car l)) (cons (car l) (flatlist (cdr l))))
((append (flatlist (car l)) (flatlist (cdr l))))
)
)

Joe Burke