List combination with common attribute

[Iraklis2009]

Hello everybody,

From what I'll say you'll understand that I have the solution right before my eyes... I am so so stuck right now.

I have two lists with lists inside:

((A) (A) (A) (B) (B) (B) (B) (C) (C))

and

((A 1) (A 2) (A 3) (B 1) (B 2) (B 3) (B 4) (C 1) (C 2))

I just want to make a list with lists that is:

((A 1 2 3) (B 1 2 3 4) (C 1 2))

I am guessing mapcar and lambda should come into play but i can't figure it out.

Any solution is more than welcome.

Thank you.

[Lee_Mac · Accepted solution]

Perhaps a 'GroupByAssoc' function:

(defun LM:GroupByAssoc ( l / a r )
    (while l
        (setq r
            (if (setq a (assoc (caar l) r))
                (subst (cons (car a) (append (cdr a) (cdar l))) a r)
                (cons  (car l) r)
            )
            l (cdr l)
        )
    )
    (reverse r)
)

Example:

_$ (LM:GroupByAssoc '((A 1) (A 2) (A 3) (B 1) (B 2) (B 3) (B 4) (C 1) (C 2)))
((A 1 2 3) (B 1 2 3 4) (C 1 2))

[Iraklis2009]

Dear Lee_Mac,

Kudos to you.

Worked like a charm.

A mile thanks.

Just a question.

Is there any reason I should reverse the list? - apart from having it in the original order.

[Lee_Mac]

You're very welcome!

---

Iraklis2009 wrote:

Just a question.

Is there any reason I should reverse the list? - apart from having it in the original order.

---

I assume you refer to the final expression in my function: (reverse r)?

This is to return the result in the correct order, since the list 'r' is constructed in reverse using the cons function. The reverse / cons combination is used since it is far faster than append to achieve the same result.

Lee