在球拍中转置矩阵(列表列表) [英] Transpose a matrix in racket (list of lists

问题描述

我在球拍中得到了一个清单列表，必须进行转置.

I got a list of lists in racket and have to transpose them.

(: transpose ((list-of(list-of %a)) -> (list-of (list-of %a))))

(check-expect (transpose (list (list 1 2 3)
                               (list 4 5 6)))
              (list (list 1 4)
                    (list 2 5)
                    (list 3 6)))

(define transpose
  (lambda (xs)
    (cond
      ((empty? xs)empty)
      ((pair? xs)(make-pair  (make-pair (first(first xs))  (make-pair (first(first(rest xs)))empty)) (transpose (rest(rest xs))))))))

这是我目前的代码. 我认为问题出在递归调用中(如果我输入错了，请纠正我).

That's my code at the moment. I think the problem is in the recursive call (correct me if I'm wrong please).

实际结果是(list (list 1 4)).其余的似乎都被忽略了.

The actual outcome is (list (list 1 4)). The rest seems kinda ignored.

如果有人知道问题或有建议，它将对我有真正的帮助.

It would really help me, if somebody knows the problem, or has a tip.

推荐答案

transpose的最简单定义是:

(define (transpose xss)
  (apply map list xss))

为什么起作用?

  (apply map list '((a b) (d e))
= (apply map List '((a b) (d e))    ; use List rather than list
= (map List '(a b) '(d e))
= (list (List 'a 'd) (List 'b e))
= '((a d) (b e))

此处List用大写字母拼写，只是为了表明哪个list是由用户给出的，而哪个是map产生的.

Here List is spelled with capital letters only to show which list was given by the user and which was produced by map.

这是一个不太聪明"的解决方案.它使用 矩阵成为转置矩阵的第一行.

Here is a less "clever" solution. It uses that the first column of a matrix becomes the first row in the transposed matrix.

(define transpose
  (lambda (xss)
    (cond
      [(empty? xss)         empty]
      [(empty? (first xss)) empty]
      [else                 (define first-column   (map first xss))
                            (define other-columns  (map rest  xss))
                            (cons first-column
                                  (transpose other-columns))])))

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