Scheme 方案：在列表和子列表中搜索元素

我想写一个函数，它接收列表并返回每个元素的列表。 例如： 获取-

（x345（x4）3x6））

并接收：

（x（x）x（））

我的代码结果用于：

(lookForX '(x 3 4 5 (x 4) 3 x (6))) 
-> (x x) 

---

我做错了什么？

在您的函数中，您只在列表中查找

x

作为元素，而没有执行子列表：

(define (filter-x lst)
  (cond
    ((null? lst) '())
    ((eq? (car lst) 'x)
     (cons (car lst)
           (filter-x (cdr lst))))
    ((pair? (car lst))
     (cons (filter-x (car lst))
           (filter-x (cdr lst))))
    (else (filter-x (cdr lst)))))

(filter-x '(x 3 4 5 (x 4) 3 x (6)))
; ==> (x (x) x ())

请注意，我将其重命名为更像lisp。Lisp代码通常不使用camelCase，而是使用Lisp case。您可以更一般地执行此操作：

(define (filter-tree predicate? lst)
  (cond
    ((null? lst) '())
    ((predicate? (car lst))
     (cons (car lst)
           (filter-tree predicate? (cdr lst))))
    ((pair? (car lst))
     (cons (filter-tree predicate? (car lst))
           (filter-tree predicate? (cdr lst))))
    (else (filter-tree predicate? (cdr lst)))))

(define (filter-tree-x lst)
  (filter-tree (lambda (v) (eq? v 'x)) lst))

(filter-tree-x '(x 3 4 5 (x 4) 3 x (6)))
; ==> (x (x) x ())

(define (filter-tree-numbers lst)
  (filter-tree number? lst))

(filter-tree-numbers '(x 3 4 5 (x 4) 3 x (6)))
; ==> (3 4 5 (4) 3 (6))

(define (filter-tree predicate? lst)
  (cond
    ((null? lst) '())
    ((predicate? (car lst))
     (cons (car lst)
           (filter-tree predicate? (cdr lst))))
    ((pair? (car lst))
     (cons (filter-tree predicate? (car lst))
           (filter-tree predicate? (cdr lst))))
    (else (filter-tree predicate? (cdr lst)))))

(define (filter-tree-x lst)
  (filter-tree (lambda (v) (eq? v 'x)) lst))

(filter-tree-x '(x 3 4 5 (x 4) 3 x (6)))
; ==> (x (x) x ())

(define (filter-tree-numbers lst)
  (filter-tree number? lst))

(filter-tree-numbers '(x 3 4 5 (x 4) 3 x (6)))
; ==> (3 4 5 (4) 3 (6))