将图形更改为Haskell中的节点列表 [英] Changing a graph to list of nodes in Haskell

问题描述

我有这些数据类型：

 数据节点a =节点
 {label :: a，
 adjacent :: [（a，Int）]}派生Show 
 data Network a = Graph [Node a]派生Show 
  

我想将图形转换为节点列表。例如，我想打开它：

$ $ $ $ $ $ $ $ $ $ c'，2）]），（Node'b'[（'c'，3）]），（Node'c'[]）]

改为：

  [（Node'a'[（'b' ，（'c'，2）]），（Node'b'[（'c'，3）]），（Node'c'[]）] 
  
 
 
 
我写了这个函数和其他一些变体：
 
 
  deGraph图形[节点xy] = [节点xy] 
  
但是我不断收到错误。你能告诉我如何改变我的功能吗？ 
谢谢。
解决方案
 
 
 
  foo [x] = x 
  
匹配单个元素的列表并将该元素绑定到x。
 
 
 
因为您希望它匹配所有列表，所以您可以执行类似于
  foo xs = xs 
  
所以你的代码应该改成 
 
 
  deGraph（图形节点）=节点
  - 注意事实我包装了构造函数
  - 在parens中
  
总结：
 
 
 
只要明确一点，以下是您可以在列表中匹配的不同方式 
 
 
  - 匹配单个元素（这是句法含义）
 foo [x，y] = x 
  - 抓取列表的头部和尾部（这是实际的解构）
 foo （x：rest）= x 
  - 匹配一个空列表
 foo [] =错误Oh noes
 - 匹配一切
 foo xs =头部xs 
  
或者上述的任意组合。 / p> 
I have these data types : 
data Node a = Node
    { label :: a,
        adjacent :: [(a,Int)] } deriving Show
data Network a = Graph [Node a] deriving Show
I want to turn a graph to a list of nodes. For example I want to turn this : 
Graph [ ( Node 'a' [ ( 'b' , 3 ) , ( 'c' ,2 ) ] ) , ( Node 'b' [ ('c' , 3 ) ] ) , ( Node 'c' [] ) ]
to this : 
[ ( Node 'a' [ ( 'b' , 3 ) , ( 'c' ,2 ) ] ) , ( Node 'b' [ ('c' , 3 ) ] ) , ( Node 'c' [] ) ]
I wrote this function and some other variations of it : 
deGraph Graph [Node x y] = [Node x y]
but I kept getting erros. Can you tell me how I should change my function?
Thanks.
解决方案
You're misunderstanding how to pattern match on a list.
foo [x] = x
matches a list of a single element and binds that element to x.


Since you want it to match on all lists, you'd do something like
foo xs = xs
so your code should change to
deGraph (Graph nodes) = nodes
-- Notice the fact that I wrapped the constructor
-- in parens
Wrap up:

Just to be explicit, here are the different ways you can match on a list
 -- matches on individual elements (this is syntactic sugary goodness)
foo [x, y] = x
 -- grabs the head and tail of the list (This is actual deconstructing)
foo (x:rest) = x
 -- matches an empty list
foo []       = error "Oh noes"
 -- matches everything
foo xs       = head xs
Or any combination of the above.

                        
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