Question

我有这段代码：

type Matrice = [[String]]
matr =[[" - "," 0 "," - "],[" - "," - "," - "],[" - "," - "," - "]]       

changeValue :: Matrice ->Int->Int->Matrice
changeValue mat x y = [
    if ((mat !! x) !! y) /= " - "
        then mat
        else do (replaceNth y " P " xs)
    | xs <- (mat !! x)
    ]

replaceNth :: function用（'P'）

替换位置值' - '

replaceNth :: Int -> String -> [String] -> [String]
replaceNth n newVal (x:xs)
     | n == 0 = newVal:xs
     | otherwise = x:replaceNth (n-1) newVal xs

我想改变每个案例在矩阵中使用' - '到'P'

但它没有用，我总是这个错误：

couldn't match type [char] with char

Answer 1

我是Haskell的新手，所以这可能在很多方面都是次优的，但我发现使用Haskell功能和多态来替换你的replaceNth函数的事实很有意思，用一个更普遍的替换元素将函数应用于所述元素的结果列表：

changeNth :: Int->(a->a)->[a]->[a]
changeNth n change (x:xs)
     | n == 0 = (change x):xs
     | otherwise = x:changeNth (n-1) change xs

现在您可以使用它两次来完成您想要的任务：

changeValue :: Matrice ->Int->Int->Matrice
changeValue mat x y = changeNth x (changeNth y
  (\v -> if v==" - " then " P " else v)) mat

我的测试：

λ: let matr =[[" - "," 0 "," - "],[" - "," - "," - "],[" - "," - "," - "]]
λ: changeValue matr 1 1
[[" - "," 0 "," - "],[" - "," P "," - "],[" - "," - "," - "]]
λ: changeValue matr 0 1
[[" - "," 0 "," - "],[" - "," - "," - "],[" - "," - "," - "]]

替换列表列表中的元素 - haskell

1 个答案: