问题
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) '()))
(Full code: https://gist.github.com/anonymous/b8cfcb0bf021be9ef9c8)
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).
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.
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.
How can I filter out the nulls of the returned list?
回答1:
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.
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)))
...)
Now, use append-map:
(append-map prime-pairs primes-list-split)
and you should have the results you seek. (See my forked gist for full code.)
来源：https://stackoverflow.com/questions/31909685/how-can-i-filter-null-values-from-this-list