定义elisp函数，获取2个列表并返回列表1中的原子出现在列表2中的次数

Question

我如何定义一个接受2个列表（l1 l2）的函数，并返回列表1中列表1中的原子出现的次数。

Answer 1

诀窍是迭代第二个列表，计算你遇到的事物中有多少出现在第一个列表中。 member函数允许您进行该测试，因此您最终可能会选择以下两个选项之一：

;; 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有一个sum函数来对列表或某种fold求和，另一种选择是在第二个列表中映射以得到0和1然后然后将它们压缩。我不认为它确实如此，所以我建议count-known-atoms-1。