如何从Scheme中的列表中删除非重复元素? [英] How to remove non-duplicate elements from a list in Scheme?

问题描述

给定一个列表，

(define ll '(a a a b c c c d e e e e))

我想删除所有非重复元素，只留下重复元素的一个副本，即删除后，结果是

I want to remove all non-duplicate elements and leave only one copy of the duplicate one, i.e. after removing, the result would be

(a c e)

我的算法是:

• 遍历列表，比较当前元素和下一个元素.

• Traverse through the list, comparing current element with next element.

• 如果它们相等，则cons 当前元素与下一个递归调用的列表.例如，

• If they're equal, then cons the current element with the list of the next recursive call. For example,

(a a a b c)

从左向右移动，遇到a和a.

Move from left to right, encounter a and a.

(cons a (remove-nondup (cddr lst)))

否则，跳过当前和下一个元素.

Otherwise, skip current and next element.

(remove-nondup (cddr lst))

我遇到的问题是

(define (remove-nondup lst)
  (if (>= (length lst) 2)
      (if (eq? (car lst) (cadr lst))
          (cons (car lst) (remove-nondup (cdr lst)))
          (remove-nondup (cddr lst)))
      lst))

我遇到的问题是，如果连续元素超过 3 个，我将无法跟踪前一个前一个.所以我想知道我是否应该使用另一个程序来删除所有重复项?或者我可以把它们放在一个程序中?

The problem that I'm having is if there are more than 3 consecutive elements, I have no way to keep track of the previous-previous one. So I wonder should I use another procedure to remove all duplicates? or I can just put them into one procedure?

所以我目前的替代解决方案是，

So my alternative current solution was,

(define (remove-dup lst)
  (if (>= (length lst) 2)
      (if (eq? (car lst) (cadr lst))
          (cons (car lst) (remove-dup (cddr lst)))
          (cons (car lst) (remove-dup (cdr lst))))
      lst))

(define (remove-nondup-helper lst)
  (if (>= (length lst) 2)
      (if (eq? (car lst) (cadr lst))
          (cons (car lst) (remove-nondup-helper (cdr lst)))
          (remove-nondup (cddr lst)))
      lst))

; call the helper function and remove-dup
(define (remove-nondup lst)
  (remove-dup (remove-nondup-helper lst)))

推荐答案

这是我的解决方案:首先，抓取 bagify(任何版本都可以).然后:

Here's my solution: first, grab bagify (any version will do). Then:

(define (remove-singletons lst)
  (define (singleton? ass)
    (< (cdr ass) 2))
  (map car (remove singleton? (bagify lst))))

remove 来自 SRFI 1.如果您使用 Racket，请先运行 (require srfi/1).或者，使用这个简单的定义:

remove is from SRFI 1. If you're using Racket, run (require srfi/1) first. Or, use this simple definition:

(define remove #f)   ; Only needed in Racket's REPL
(define (remove pred lst)
  (cond ((null? lst) lst)
        ((pred (car lst)) (remove pred (cdr lst)))
        (else (cons (car lst) (remove pred (cdr lst))))))

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