计数物品的优雅方式 [英] elegant way to count items

问题描述

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:

How can I count the total number of occurrences of the terms in the car of each element in the list? Basically I want:

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

不，这不是家庭作业.我只是还没有在 Lisp 中思考"的东西.

No, this is not homework. I just don't have the "thinking in Lisp" thing quite yet.

在 C# 中，我将使用 LINQ 来执行此操作.我也可以在 lisp 中做到这一点，使用 while 循环等，但我想这样做的方式似乎过于复杂.

In C#, I would use LINQ to do this. I can do it in lisp, too, using while loops and such but the way I am thinking of doing it seems overly complicated.

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))))

推荐答案

我不知道这是最优雅的，但似乎合理:

I dunno that this is the most elegant, but it seems reasonable:

(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))))))

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