haskell - 替换列表列表中的元素——haskell

Question

我有这个代码：

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 :: 函数将位置值 ' - ' 替换为 (' P ')

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

我想将矩阵中的每个案例都更改为 ' - ' 为 ' P '

但它不工作，我总是这个错误：

couldn't match type [char] with char

Answer 1

我是 Haskell 的新手，所以这在很多方面可能不是最优的，但我发现使用 Haskell 是函数式和多态的这一事实来用更通用的替换你的 replaceNth 函数很有趣，它替换列表中的元素将函数应用于所述元素的结果：

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

现在你可以使用它两次来完成你想要的：

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 "," - "],[" - "," - "," - "],[" - "," - "," - "]]