将[String]列表转换为Haskell中的[Double]列表

Question

我正在写一个函数来将字符串列表（从CSV文件读取）转换为双精度列表。它在第三行给我一个错误。

stringToDouble :: [String] -> [Double] 
stringToDouble [] = error "empty list" 
stringToDouble [x] = read x :: Double -- the `read` gives me an error 
stringToDouble (x:xs) = stringToDouble xs 

是否因为我没有将转换后的Double放入需要返回的列表中？

Answer 1

错误来自于read x类型为Double而不是[Double]，但是就其本身而言，即使使用该修补程序，您的功能也无法正常工作。

让我们把你的函数放到单词中：“把字符串列表的前面的元素作为一个double来读取，然后对列表的其余部分做同样的处理”。现在让我们来看看你的功能：

stringToDouble :: [String] -> [Double] 
stringToDouble [] = error "empty list" 
stringToDouble [x] = read x :: Double -- Error 
stringToDouble (x:xs) = stringToDouble xs 

现在让我们将修复应用到它。另外，没有理由在空列表中出错。只是产量和空的双打名单：

stringToDouble :: [String] -> [Double] 
stringToDouble [] = [] 
stringToDouble [x] = [read x :: Double] -- Put the single value into a list 
stringToDouble (x:xs) = stringToDouble xs 

问题在于递归步骤。在列表上调用stringToDouble与在列表的尾部调用stringToDouble相同。第一个元素简单地被丢弃。您想要转换头并将其放回列表中。

stringToDouble :: [String] -> [Double] 
stringToDouble [] = [] 
stringToDouble [x] = [read x :: Double] -- Put the single value into a list 
stringToDouble (x:xs) = (read x :: Double) : stringToDouble xs 

哪里(:)是用于将元件连接到列表中的正面的操作员。因此，甚至不需要中间线，因为递归步骤将处理转换，空列表步骤将处理停止条件。

stringToDouble :: [String] -> [Double] 
stringToDouble [] = [] 
stringToDouble (x:xs) = (read x :: Double) : stringToDouble xs 

现在，实事求是，你可以可能删除:: Double部分和Haskell会找出你的意思与功能的类型约束，但它不会伤害，有时它有助于可读性离开它英寸

Answer 2

你是对的 - read x :: Double是Double类型，而函数的返回类型是[Double]（意思是“Double”的列表）。

事情是这样的：

stringToDouble [x] = [read x :: Double] 

应该工作。

请注意，如果您尝试单独转换列表中的每个元素，则应该使用map而不是使用显式递归来编写函数。如果f是String -> Double类型，则map f将是[String] -> [Double]类型的函数。