Scheme：在列表和子列表中搜索元素

Question

我想写一个接收列表并返回每个元素列表的函数。 例如： GET - (x 3 4 5 (x 4) 3 x (6)))和接收：

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

我在做什么错：为(x (x) x())

(define (lookForX lst) 
    (cond 
    ((null? lst) '()) 
    ((eq? (car lst) 'x) (cons (car lst) (lookForX (cdr lst)))) 
    (else (lookForX (cdr lst))))) 

我的代码的结果？

Answer 1

在你工作，你只是在寻找x为列表元素，你是不是做子列表：

(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()) 

通知我改名这更像口齿不清。 Lisp代码通常不使用camelCase，而是使用lisp-case。你可以做得更一般：

(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))