如何从此列表中过滤空值? [英] How can I filter null values from this list?

问题描述

我按照以下步骤在列表中创建所有素数对:

I have the following procedure for creating all prime-pairs in a list:

(define (prime-pairs lst)
  (define (split lst pos)
    (list (drop-right lst pos) (take-right lst pos)))
  (define (prime-pairs-iter n acc)
    (cond ((= n 0) (filter (lambda (e) (not (null? e))) acc))
          (else (prime-pairs-iter (- n 1) 
                                  (let ((s (split lst n)))
                                    (if (and (prime? (list->number (car s)))
                                             (prime? (list->number (cadr s))))
                                        (append s acc)
                                        acc))))))
  (prime-pairs-iter (- (length lst) 1) '()))

(完整代码: https://gist.github.com/anonymous/b8cfcb0bf021be9ef9c8 )

我要prime-pairs要做的是在lst中创建由两个素数组成的每对列表.数字以列表格式表示，如下所示:11将为'(1 1).

What I want prime-pairs to do is create a list of every pair in the lst that consists of two primes. The numbers are represented in a list format as follows: 11 would be '(1 1).

不幸的是，当我运行这段代码时，(filter (lambda (e) (not (null? e))) acc))似乎并没有从最终结果中删除'()，最终我得到了一长串空值和想要的对.

Unfortunately, when I run this code the (filter (lambda (e) (not (null? e))) acc)) does not seem to remove the '() from the end result, and I end up with a long list of empty values and the wanted pairs.

如果我使用(filter null? acc))，则确实会保留一个空值列表.因此，另一种方法(过滤出实际值)确实可以工作.

If I use a (filter null? acc)) a list of empty values does remain. So the other way around (filtering out the actual values) does work.

如何过滤返回列表的空值?

How can I filter out the nulls of the returned list?

推荐答案

当前，您的prime-pairs函数始终返回一个值:空列表或素数对.使用map，如果不对map的结果进行进一步的过滤，则无法避免空列表.

Currently, your prime-pairs function always returns one value: either an empty list, or the prime pairs. Using map, there is no way to avoid the empty lists, without doing further filtering on the result of the map.

一种替代方法是返回结果的列表，并使用append-map而不是map.更改您的prime-pairs以返回一个空列表，或者返回一个包含素数对的单例列表；这模拟返回零或一个值，而不总是返回一个值.像这样:

One alternative is to return a list of results, and use append-map instead of map. Change your prime-pairs to return either an empty list, or a singleton list containing your prime pairs; this simulates returning zero or one value, rather than always one value. Like so:

(cond ((zero? n) (if (null? acc)
                     '()
                     (list acc)))
      ...)

现在，使用append-map:

(append-map prime-pairs primes-list-split)

，您应该拥有想要的结果. (有关完整代码，请参见我的分叉要领.)

and you should have the results you seek. (See my forked gist for full code.)

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