用空列表填充元组列表 [英] Transpose list of tuples filling with empty lists

问题描述

我是Scheme的新手，我正在尝试编写一个将 n 列表组合成 n -tuple列表的过程.如果列表的大小不同，则当相应列表用完元素时，元组应包含空列表().

I'm new to Scheme and I'm trying to write a procedure which combines n list into a list of n-tuples. If the lists are of different size, the tuples should contain the empty list () when the corresponding list ran out of elements.

我当前的实现如下:

(define (comb list1 list2)  
    (cond [(empty? list1) empty]
          [(empty? list2) empty]
          [else (cons (list (first list1) (first list2))
                      (comb (rest list1) (rest list2)))]))

但是，当列表中没有更多项目可合并时，此程序不会生成另一个元组.例如，(comb '(1 2 3 ) '(3 4))仅产生((1 3) (2 4))

However, this program doesn't produce another tuple when there are no more items in the list to combine. For instance, (comb '(1 2 3 ) '(3 4)) produces only ((1 3) (2 4))

我该如何解决?

推荐答案

这有点棘手，我认为这对于只学习该语言基础知识的人来说并不适合.无论如何，这是我针对高阶程序提出的解决方案:

This is a bit tricky, and I believe it's not an appropriate exercise for someone who is just learning the basics of the language. Anyway, here's my proposed solution, in terms of higher-order procedures:

; helper procedure for filling a list with arbitrary values at the end
(define (fill lst val num)
  (append lst
          (build-list num (const val))))

; helper procedure for transposing a list of lists    
(define (transpose lsts)
  (apply map list lsts))

; main procedure
(define (list-tuples lsts)
  (let* ((lengths    (map length lsts))    ; obtain the length of each sublist
         (max-length (apply max lengths))) ; find out the maximum length
    (transpose                             ; build new sublists element-wise
     (map (lambda (lst len)                ; build sublists of the right length
            (fill lst '() (- max-length len)))  ; fill sublists with '()
          lsts
          lengths))))

诀窍是找到列表的最大长度，然后以该长度构建新列表，最后以'()填充它们.之后，只需从每个子列表中选取一个元素即可构建答案.它可以按预期工作:

The trick was to find the maximum length of the lists and then build new lists with that length, filling them with '() at the end. After that, it's a simple matter of building the answer by taking one element from each sublist. It works as expected:

(list-tuples '((m n o) (1) (x y)))
=> '((m 1 x) (n () y) (o () ()))

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