Haskell Data.Graph graphFromEdges 示例

Question

我目前正在尝试为 Haskell 编写一个 GraphML 阅读器(请参阅 this previous question )。由于我对 Haskell 还很陌生，尽管学习曲线陡峭，但我还没有完全学会仅从 Hackage 文档中推导出示例的能力。

我当前拥有的是一个 Graph 实例:

data Graph = Graph
  { graphId :: String,
    nodes :: [String],
    edges :: [(String, String)] -- (Source, target)
  }
  deriving (Show, Eq)

我想将它们转换为 Data.Graph 。我认为 graphFromEdges 或 graphFromEdges' 是执行此操作的正确函数。

我当前的尝试是创建三元组(nodeid，nodeid，[边缘目标])

-- Convert a graph node into a Data.Graph-usable
getDataGraphNode :: Graph -> String -> (String, String, [String])
getDataGraphNode graph node = (node, node, getTargets node graph)

-- Convert a Graph instance into a Data.Graph list of (node, nodeid, edge) tuples
getDataGraphNodeList :: Graph -> [(String, String, [String])]
getDataGraphNodeList graph = map (getDataGraphNode graph) (nodes graph)

但是，由于类型错误，GHC 不会编译它:

 Couldn't match expected type `[(node0, key0, [key0])]'
            with actual type `(String, String, [String])'

您能否给我指出一个示例，或者最好描述一下如何从文档中推导出示例的一般方法(在本例中为函数签名)？我是否需要使用 Data.Graph 中的 Vertex 类型？我目前无法弄清楚 node0 和 key0 类型需要是什么。

这是一个最小的示例，显示了我在使用 Data.Graph 时遇到的问题。我不确定这与误用类型有何关系，即使我自己的代码中的许多错误是由于类型系统问题而发生的:

import Data.Graph
main = do
    graph <- graphFromEdges' [("n0","n0",["n1"]), ("n1","n1",["n2"]), ("n2","n2",[])]
    -- This should print ["n0","n1","n2"]
    print $ vertices graph

它会产生以下错误消息:

bar.hs:4:5:
Couldn't match type `IO' with `(,) Graph'
    Expected type: (Graph, ())
      Actual type: IO ()
    In a stmt of a 'do' block: print $ vertices graph
    In the expression:
      do { graph <- graphFromEdges'
                      [("n0", "n0", ["n1"]), ("n1", "n1", ["n2"]), ....];
           print $ vertices graph }
    In an equation for `main':
        main
          = do { graph <- graphFromEdges' [("n0", "n0", [...]), ....];
                 print $ vertices graph }

bar.hs:4:22:
    Couldn't match type `Vertex -> ([Char], [Char], [[Char]])'
                  with `GHC.Arr.Array Vertex [Vertex]'
    Expected type: Graph
      Actual type: Vertex -> ([Char], [Char], [[Char]])
    In the first argument of `vertices', namely `graph'
    In the second argument of `($)', namely `vertices graph'
    In a stmt of a 'do' block: print $ vertices graph

Answer 1

试试这个:

import Data.Graph
main = do
    let (graph, vertexMap) = graphFromEdges' [("n0","n0",["n1"]), ("n1","n1",["n2"]),("n2","n2",[])]
    -- This should print ["n0","n1","n2"]
    print $ map ((\ (vid, _, _) -> vid) . vertexMap) (vertices graph)

graphFromEdges' 返回一对值和 Graph是那对中的第一个。另外，在这种情况下<-用于返回 IO something 的东西而graphFromEdges'返回一个纯值。

关键错误是这样的:

Couldn't match type `IO' with `(,) Graph'
    Expected type: (Graph, ())
      Actual type: IO ()

虽然报告的方式有点误导 - 如果您给出main，报告会更好类型签名 main :: IO () 。一般来说，如果您对类型错误感到困惑，那么值得尝试找出您认为事物应该是什么类型并使用显式类型签名。