使用合并排序的方案向量 [英] Scheme Vector using merge sorting

问题描述

我有一个向量，每个向量的元素都是一个列表，我想根据列表的长度对元素进行排序.我正在用它来对向量进行排序，但是出现了错误

I have a vector, the elements of each vector is a list, I want to sort the elements regarding to the length of list. I am using this to sort my vector but I got the error

    (define vector-merge!
   (lambda (newvec vec left group-size vec-size)
      (let* ((top-left (min vec-size (+ left group-size)))
             (right top-left)
             (top-right (min vec-size (+ right group-size))))
         (let mergeloop ((left left) (right right) (i left))
              (cond ((and (< left top-left) (< right top-right))
                        (if (< (vector-ref vec left) (vector-ref vec right))
                       (begin
                          (vector-set! newvec i (vector-ref vec left)) 
                          (mergeloop (add1 left) right (add1 i)))
                       (begin
                          (vector-set! newvec i (vector-ref vec right)) 
                          (mergeloop left (add1 right) (add1 i)))))
                ((< left top-left)
                    (vector-set! newvec i (vector-ref vec left))
                    (mergeloop (add1 left) right (add1 i)))
                ((< right top-right)
                    (vector-set! newvec i (vector-ref vec right))
                    (mergeloop left (add1 right) (add1 i))))))))

  (define vector-mergesort!
   (lambda (orig-vec)
      (let* ((vec-size (vector-length orig-vec))
             (new-vec (make-vector vec-size)))
        ;; merge with successively larger group sizes
        (do ((group-size 1 (* group-size 2))    ;; loop variables
         (twice-size 2 (* twice-size 2))
         (count 1 (add1 count))
         (vec1 orig-vec vec2)
         (vec2 new-vec vec1))
        ((>= group-size vec-size)          ;;; exit condition
            (if (even? count)              ;;; copy to orig-vec, if needed
                    (do ((i 0 (add1 i)))   ;;; this do replaces 
                        ((>= i vec-size))  ;;; vector-change!
                        (vector-set! orig-vec i (vector-ref new-vec i)))))
        ;; successively merge next two groups
        (do ((left 0 (+ left twice-size)))    ;; loop variables
            ((>= left vec-size))              ;; exit when array processed
            (vector-merge! vec2 vec1 left group-size vec-size))))))



Error:
<: expects type <real number> as 1st argument, given: ((length (vector-ref route number))); other arguments were: ((length (vector-ref route number)))

推荐答案

这是表示错误的表达式:

This is the expression that signals an error:

(< (vector-ref vec left) (vector-ref vec right))

功能<期望将实数作为第一个参数，但得到一个列表. 由于您的vector vec包含列表，因此表达式(vector-ref vec左) 返回列表(而不是数字).由于您想按长度排序 列表中，您需要编写:

The function < expects a real number as a first argument, but got a list. Since your vector vec contains lists, the expression (vector-ref vec left) returns a list (and not a number). Since you want to sort after the length of the lists, you need to write:

(< (length (vector-ref vec left)) (length (vector-ref vec right)))

为了比较列表的长度而不是列表本身.

in order to compare the length of the lists instead of the lists themselves.

注意:您的Scheme实现很可能在其库中具有向量排序功能.在R6RS中，该过程称为向量排序！:

Note: Your Scheme implementation most probably has a vector sort function in its library. In R6RS the procedure is called vector-sort!:

(vector-sort! proc vector) 

其中proc是用于比较两个元素的过程，而vector是要排序的向量.

where proc is a procedure used to compare two elements and vector is the vector to be sorted.

因此，如果您定义:

(define (compare list1 list2)
    (< (length list1) (length list2)))

您可以对它进行排序

(vector-sort! compare vector)

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