Topic: (challenge)? ASSOCiate by value vs. key

[daron]

Take this list:

Code: [Select]

(setq stdWallList '((0 . "Basement") (1 . "Garage") (2 . "Stoop")
    (3 . "Arm") (4 . "Haunch")
    )
  )
and find the associated value of "Stoop".

What I'm looking for is the reverse of:

Code: [Select]

(cdr (assoc 2 stdWallList)

[CAB]

Code: [Select]

(cadr (assoc "Stoop" (mapcar '(lambda (x) (list (cdr x)(car x))) stdWallList)))
edit - changed to cadr

[daron]

That's good CAB. I came up with:

Code: [Select]

(defun revassoc (itempred itemcomp / newitemlist)
     (foreach item itempred
 (if (= itemcomp (cdr item))
      (setq newitemlist (append newitemlist (list item)))
 )
     )
     newitemlist
)
Not too challenging. I'd like to see the varied possibilities.

[whdjr]

Code: [Select]

(defun test (str lst)
  (caar (vl-remove-if-not '(lambda (x) (eq (cdr x) str)) lst))
)

[CAB]

Thanks,
I changed mine to cadr ILO cdr

[daron]

With a small edit, the one I made above can be made to reflect exactly the way assoc does when it encounters multiple key values:

Code: [Select]

;example: (revassoc stdwalllist "Garage")
(defun revassoc (itempred itemcomp / newitemlist)
     (foreach item itempred
 (if (= itemcomp (cdr item))
      (setq newitemlist (append newitemlist (list item)))
 )
     )
     (car newitemlist)
)
My previous one returns a list of all key's equal. This one returns the first of all in just a list, instead of a nested list.

[CAB]

Here is another.

Code: [Select]

(defun revassoc (itm lst)
  (vl-some '(lambda (x) (if (= (cdr x) itm)(car x))) lst)
)

Did you want the lisy back like this?

Code: [Select]

(defun revassoc (itm lst)
  (vl-some '(lambda (x) (if (= (cdr x) itm) x)) lst)
)

[Jeff_M]

Code: [Select]

(cadr (assoc "Stoop" (mapcar '(lambda (x) (list (cdr x)(car x))) stdWallList)))
edit - changed to cadr

I would keep the use of (cdr) since you are supposed to be using an association list. Which means you need to build the association list in reverse....just a slight mod. to CAB's code:

Code: [Select]

(cdr (assoc "Stoop" (mapcar '(lambda (x) (cons (cdr x)(car x))) stdWallList)))  

[CAB]

What we have here is a failure to communicate.

What I'm looking for is the reverse of:

Code: [Select]

(cdr (assoc 2 stdWallList))

That doesn't return a list.  

[Jeff_M]

That doesn't return a list.  

Nor does the code I modified with (cons) in place of (list)    

[CAB]

Oh, I see what you did.  The internal result is a dotted pair.
That way it has the same look as the usual code.

How about this:

Code: [Select]

(defun rv_list (lst)
  (mapcar '(lambda (x) (cons (cdr x)(car x))) lst)
)


(cdr (assoc "Stoop" (rv_list stdWallList)))

[Jeff_M]

Exactly! I have one of these for switching around layer names for a client with a strange layer naming "standard". The layers are in an assoc list ("MyLayer" . "HisLayer"), etc., then when I want to flip to my layers I just call it (fliplay laylist nil) or to his (fliplay laylist t)....depending the arg. being true I just flip 'em around.

[daron]

Sorry if there was any miscom. I was trying to leave it a little open. It's good for variations that don't get too far away from the point. It could spur ideas that weren't there to begin with. I should've left out the cdr assoc bit.

[SMadsen]

Code: [Select]

(defun rvAssoc (item lst)
  (cond ((null lst) nil)
        ((eq (cdar lst) item) (car lst))
        ((rvAssoc item (cdr lst)))
  )
)

[daron]

Sweet recursion. I had hoped for something like that too. Thanks Stig.