此转置功能如何工作? [英] How does this transpose function work?

问题描述

我的教授提供了该功能，我不知道实际发生了什么.

I have this function given by my professor and I have no idea what is actually happening.

这里是计算m×n矩阵的转置的函数:

Here is the function that computes the transpose of an m-by-n matrix:

let rec transpose = function
| [] -> failwith "cannot transpose a 0-by-n matrix"
| []::xs -> [] 
| xs -> List.map List.head xs :: transpose (List.map List.tail xs)

测试功能:

> transpose [[1;2;3];[4;5;6]];;
  val it : int list list = [[1; 4]; [2; 5]; [3; 6]]

我了解List.map，递归和所有其他内容.我只是不明白为什么/如何使用此功能.任何澄清将不胜感激！谢谢！

I understand List.map, recursion and all that stuff. I just do not understand why/how this function works. Any clarifications will be much appreciated! Thanks!

推荐答案

假设我们有一个3x3矩阵A.

Say we have a 3x3 matrix A.

let A = 
  [ [1;2;3]
    [4;5;6]
    [7;8;9] ]

= [ R1
    R2
    R3 ]

现在，让我们分析一下转置函数.

Now, let's analyse the transpose function.

let rec transpose = function
| [] -> failwith "cannot transpose a 0-by-n matrix"
| []::xs -> [] 
| xs -> List.map List.head xs :: transpose (List.map List.tail xs)

前2个案例捕获了以下事件:

The first 2 cases catch the events where:

• 该列表为空列表
• 该列表包含一个空列表，表示所有内部列表均已处理

代码[1]

List.map List.head xs

将内部列表映射到其各自的head元素

maps the inner lists to their respective head elements

R1.Head ; R2.Head ; R3.Head
= 1 ; 4 ; 7
= C1

代码[2]

transpose (List.map List.tail xs)

(递归)转置斩首列表的尾部. 因此，在每次递归中，一列都会转置为一行. 然后，使用::关键字将这些行用于构造结果列表.

(recursivly) transposes the tails of the beheaded lists. So on each recursion a column gets transposed into a row. Using the :: keyword these rows are then used to construct the resulting list.

transpose A
= C1 :: C2 :: C3 :: []
= [ C1
    C2
    C3 ]
= [ [1;4;7]
    [2;5;8]
    [3;6;9] ]

这篇关于此转置功能如何工作?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！