返回列表的前n个 [英] returns the first n of list

问题描述

如何返回列表的前n个元素?这就是我所拥有的:

How to return the first n elements of a list? Here's what I have:

(define returns(lambda (list n)
 (cond ((null? list) '())
 (((!= (0) n) (- n 1)) (car list) (cons (car list) (returns (cdr list) n)))
        (else '()))))

示例:

(returns '(5 4 5 2 1) 2)
(5 4)

(returns '(5 4 5 2 1) 3)
(5 4 5)

推荐答案

您要的是

You're asking for the take procedure:

(define returns take)

(returns '(5 4 5 2 1) 2)
=> (5 4)

(returns '(5 4 5 2 1) 3)
=> (5 4 5)

这看起来像作业，所以我想您必须从头开始实施它.一些提示，请填写空白:

This looks like homework, so I guess you have to implement it from scratch. Some hints, fill-in the blanks:

(define returns
  (lambda (lst n)
    (if <???>                     ; if n is zero
        <???>                     ; return the empty list
        (cons <???>               ; otherwise cons the first element of the list
              (returns <???>      ; advance the recursion over the list
                       <???>))))) ; subtract 1 from n

别忘了测试！

这篇关于返回列表的前n个的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！