定义elisp函数，获取2个列表并返回列表1中原子在列表2中出现的次数

Question

我将如何定义一个接受2个列表（l1 l2）并返回列表1中原子在列表2中出现的次数的函数。

Answer 1

诀窍是遍历第二个列表，计算遇到的很多东西出现在第一个列表中。 member函数可让您进行测试，因此最终可能会遇到以下两个选项之一：

;; A version with explicit recursion down the list
;;
;; This will blow its stack if list is too long.
(defun count-known-atoms (known list)
  "Return how many of the elements of `list' are atoms and appear
  in `known'."
  (if (null list)
      0
    (let ((hd (car list)))
      (+ (if (and (atom hd) (member hd known)) 1 0)
         (count-known-atoms known (cdr list))))))

;; A version using local variables and side effects. Less pretty, if you're a
;; fan of functional programming, but probably more efficient.
(defun count-known-atoms-1 (known list)
  "Return how many of the elements of `list' are atoms and appear
  in `known'."
  (let ((count 0))
    (dolist (x list count)
      (when (and (atom x) (member x known))
        (setq count (1+ count))))))

;; (count-known-atoms   '(1 2) '(0 1 2 3 4 5))    => 2
;; (count-known-atoms-1 '(1 2) '(0 1 '(2) 3 4 5)) => 1

如果ELisp具有sum函数可对列表或某种fold求和，则另一种选择是在第二个列表上进行映射以得到零和一，然后将其压缩下来。不过，我不认为这样做，所以建议使用count-known-atoms-1。