emacs - 计算项目的优雅方式

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术语在汽车中出现的总次数？基本上我想要:

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

不，这不是家庭作业。我只是还没有“用 Lisp 思考”的东西。

在 C# 中，我会使用 LINQ 来执行此操作。我也可以用 lisp 来做，使用 while 循环等等，但我想做的方式似乎过于复杂。

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编辑

这是我的:

(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)