elegant way to count items

Question

I have a list shaped like this:

  \'((\"Alpha\" .  1538)
    (\"Beta\"  .  8036)
    (\"Gamma\" .  8990)
    (\"Beta\"  .  10052)
    (\"Alpha\" .  12837)
           


        

Answer 1

Combining higher level Common Lisp functions:

(defun count-unique (alist) 
  (mapcar
    (lambda (item)
      (cons (car item)
            (count (car item) alist :test #'equal :key #'car)))
    (remove-duplicates alist :test #'equal :key #'car)))

It doesn't scale to large lists though. If you need O(n) performance use a hash table based solution instead, such as the less elegant:

(defun count-unique (alist)
  (loop
     with hash = (make-hash-table :test #'equal)
     for (key . nil) in alist
     do (incf (gethash key hash 0))
     finally (return
               (loop for key being each hash-key of hash
                  using (hash-value value)
                  collect (cons key value)))))