将列表变成矩阵 [英] Turn a list into a matrix

问题描述

我整个下午都在做一件简单的事情，但由于某种原因似乎无法做到:如何将列表转换为给定宽度的矩阵.

I worked all afternoon on a simple thing but cannot seem to get it right for some reason : how to turn a list into a matrix of given width.

示例:我得到了一个列表，例如

Example : I got a list such as

[1, 3, 5, 7, 6, 8, 9, 0]

并想创建一个矩阵，例如

and want to create a matrix such as

[[1, 3],
 [5, 7],
 [6, 8],
 [9, 0]]

通过谓词

list2matrix/3 : list2matrix(List, Size_of_Rows, Matrix).

在这个例子中使用如下:

In this example used like :

list2matrix([1, 3, 5, 7, 6, 8, 9, 0], 2, Matrix).

如果列表的长度不是行大小的倍数，则谓词应该失败.

The predicate should fail if the length of the list is not a multiple of the size of the rows.

我决定不发布我的作品，因为我认为我做错了，无法帮助我纠正它;(

I decided not to post my work since I think I got it so wrong that it would not help me to get correction on it ;(

如果您能提出有关如何处理此类问题的任何线索，请提前致谢.

Thanks by advance if you can propose any leads about how to deal with such a problem.

推荐答案

您可以将问题分为两部分.第一个构建块是构建一行 N 个元素.即把输入列表分成两个列表，一个正好有N个元素(行)，另一个是输入列表的剩余部分.

You can divide the problem in two parts. The first building block would be to build a row of N elements. That is to take the input list and split it in two lists, one will have exactly N elements (the row) and the other is the remaining of the input list.

第二个构建块是构建由行组成的矩阵.

The second building block would be to build the matrix which is made of rows.

list_to_matrix([], _, []).
list_to_matrix(List, Size, [Row|Matrix]):-
  list_to_matrix_row(List, Size, Row, Tail),
  list_to_matrix(Tail, Size, Matrix).

list_to_matrix_row(Tail, 0, [], Tail).
list_to_matrix_row([Item|List], Size, [Item|Row], Tail):-
  NSize is Size-1,
  list_to_matrix_row(List, NSize, Row, Tail).

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