* best way to get the list of unique keys from 2 alists
@ 2005-12-16 17:47 Kevin Rodgers
2005-12-17 0:13 ` Drew Adams
2005-12-17 0:36 ` Kevin Rodgers
0 siblings, 2 replies; 6+ messages in thread
From: Kevin Rodgers @ 2005-12-16 17:47 UTC (permalink / raw)
I've got 2 alists of (SYMBOL . "STRING") pairs, and I need to get the
list of unique symbol names to pass to completing-read as its TABLE
argument: (("SYMBOL-NAME") ...)
I know about remove-duplicates and union, but I'd like to avoid using
cl*.el functions.
Since completing-read seems to ignore nil entries in TABLE, this is what
I've got now:
(nconc (mapcar (lambda (assoc)
(list (symbol-name (car assoc))))
alist-1)
(mapcar (lambda (assoc)
(or (assq (car assoc) alist-1)
(list (symbol-name (car assoc)))))
alist-2))
Is there a cleaner way?
Thanks,
--
Kevin
^ permalink raw reply [flat|nested] 6+ messages in thread
* RE: best way to get the list of unique keys from 2 alists
2005-12-16 17:47 Kevin Rodgers
@ 2005-12-17 0:13 ` Drew Adams
2005-12-17 0:36 ` Kevin Rodgers
1 sibling, 0 replies; 6+ messages in thread
From: Drew Adams @ 2005-12-17 0:13 UTC (permalink / raw)
I've got 2 alists of (SYMBOL . "STRING") pairs, and I need to get the
list of unique symbol names to pass to completing-read as its TABLE
argument: (("SYMBOL-NAME") ...)
I know about remove-duplicates and union, but I'd like to avoid using
cl*.el functions.
Since completing-read seems to ignore nil entries in TABLE, this is what
I've got now:
(nconc (mapcar (lambda (assoc)
(list (symbol-name (car assoc))))
alist-1)
(mapcar (lambda (assoc)
(or (assq (car assoc) alist-1)
(list (symbol-name (car assoc)))))
alist-2))
Is there a cleaner way?
That looks good to me.
You could always define union or remove-duplicates, but removing duplicates
would be less efficient than what you're doing (consing up the second list
completely, just to perhaps remove stuff). Anyway, here are non-destructive
versions of remove-duplicates and union, FWIW:
(defun remove-duplicates (list)
"Copy of LIST with duplicate elements removed.
Tested with `equal'."
(let ((tail list)
new)
(while tail
(unless (member (car tail) new)
(push (car tail) new))
(pop tail))
(nreverse new)))
(defun union (list1 list2)
"Combine LIST1 and LIST2 using a set-union operation.
The result list contains all items that appear in either LIST1 or
LIST2. This is a non-destructive function; it copies the data if
necessary."
(cond ((null list1) list2)
((null list2) list1)
((equal list1 list2) list1)
(t
(or (>= (length list1) (length list2))
(setq list1 (prog1 list2 (setq list2 list1)))) ; Swap them.
(while list2
(unless (member (car list2) list1)
(setq list1 (cons (car list2) list1)))
(setq list2 (cdr list2)))
list1)))
^ permalink raw reply [flat|nested] 6+ messages in thread
* Re: best way to get the list of unique keys from 2 alists
2005-12-16 17:47 Kevin Rodgers
2005-12-17 0:13 ` Drew Adams
@ 2005-12-17 0:36 ` Kevin Rodgers
1 sibling, 0 replies; 6+ messages in thread
From: Kevin Rodgers @ 2005-12-17 0:36 UTC (permalink / raw)
Kevin Rodgers wrote:
> I've got 2 alists of (SYMBOL . "STRING") pairs, and I need to get the
> list of unique symbol names to pass to completing-read as its TABLE
> argument: (("SYMBOL-NAME") ...)
>
> I know about remove-duplicates and union, but I'd like to avoid using
> cl*.el functions.
>
> Since completing-read seems to ignore nil entries in TABLE, this is what
> I've got now:
>
> (nconc (mapcar (lambda (assoc)
> (list (symbol-name (car assoc))))
> alist-1)
> (mapcar (lambda (assoc)
> (or (assq (car assoc) alist-1)
> (list (symbol-name (car assoc)))))
> alist-2))
>
> Is there a cleaner way?
Since completing-read also accepts an obarray for its TABLE, I think
this might be a little better:
(let ((table (make-vector 83 0))) ; 83 is a suitable prime
(mapc (lambda (assoc)
(intern (symbol-name (car assoc)) table))
alist-1)
(mapc (lambda (assoc)
(intern (symbol-name (car assoc)) table))
alist-2)
table)
--
Kevin Rodgers
^ permalink raw reply [flat|nested] 6+ messages in thread
end of thread, other threads:[~2005-12-20 3:35 UTC | newest]
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2005-12-18 15:20 ` best way to get the list of unique keys from 2 alists Stefan Monnier
2005-12-19 18:50 ` Kevin Rodgers
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2005-12-20 3:35 ` Stefan Monnier
2005-12-16 17:47 Kevin Rodgers
2005-12-17 0:13 ` Drew Adams
2005-12-17 0:36 ` Kevin Rodgers
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