计算物品的优雅方式

Question

我有一个这样的列表:

  '(("Alpha" .  1538)
    ("Beta"  .  8036)
    ("Gamma" .  8990)
    ("Beta"  .  10052)
    ("Alpha" .  12837)
    ("Beta"  .  13634)
    ("Beta"  .  14977)
    ("Beta"  .  15719)
    ("Alpha" .  17075)
    ("Rho"   .  18949)
    ("Gamma" .  21118)
    ("Gamma" .  26923)
    ("Alpha" .  31609))

如何计算列表中每个元素的汽车中术语的总出现次数？基本上我想要:

(("Alpha" . 4)
 ("Beta" . 5)
 ("Gamma" . 3)
 ("Rho" . 1))

不,这不是功课.我还没有"在Lisp中思考"的事情.

在C#中,我会使用LINQ来执行此操作.我也可以在lisp中使用while循环等,但我想这样做的方式似乎过于复杂.

---

编辑

这就是我所拥有的:

(defun count-uniq (list)
  "Returns an alist, each item is a cons cell where the car is
a unique element of LIST, and the cdr is the number of occurrences of that
unique element in the list. "
  (flet ((helper (list new)
                 (if (null list)
                     new
                   (let ((elt (assoc (car list) new)))
                     (helper (cdr list)
                             (if elt
                                 (progn (incf (cdr elt)) new)
                               (cons (cons (car list) 1) new)))))))
    (nreverse (helper list nil))))

Answer 1

(defun freqs (list &optional test key)
  (let ((h (make-hash-table :test test)))
    (dolist (x list)
      (let ((key (if key (funcall key x) x)))
        (puthash key (1+ (gethash key h 0)) h)))
    (let ((r nil))
      (maphash #'(lambda (k v) (push (cons k v) r)) h)
      (sort r #'(lambda (x y) (< (cdr x) (cdr y)))))))

(freqs '(("Alpha" .  1538)
         ("Beta"  .  8036)
         ("Gamma" .  8990)
         ("Beta"  .  10052)
         ("Alpha" .  12837)
         ("Beta"  .  13634)
         ("Beta"  .  14977)
         ("Beta"  .  15719)
         ("Alpha" .  17075)
         ("Rho"   .  18949)
         ("Gamma" .  21118)
         ("Gamma" .  26923)
         ("Alpha" .  31609))
       #'equal #'car)

Answer 2

我不知道这是最优雅的，但看起来很合理：

(defun add-for-cheeso (data)
  (let (result)
    (dolist (elt data result)
      (let ((sofar (assoc (car elt) result)))
        (if sofar
            (setcdr sofar (1+ (cdr sofar)))
          (push (cons (car elt) 1) result))))))