haskell 查找两个列表之间相似性的程序

Question

好的，所以我是 Haskell 的新手，我想编写一个程序，在其中获取两个列表并找出它们之间的相似性（共同 items/number 项的数量）。这是我目前所拥有的：

fun2 :: [Int]->[Int]->Float

fun2 [] xs2 = 0
fun2 xs1 xs2 = if head xs1 == head xs2 then (1/length xs2) + fun2 tail xs1 xs2
else if head xs1 /= head xs2 then fun2 xs1 tail xs2
else fun2 tail xs1 xs2

main = fun2 [1,2,3,4] [3,4,5,6]

所以让我解释一下我想做什么，我定义了我的函数来获取两个整数列表并输出一个浮点数（相似性百分比）。然后我编写我的函数，基本情况是第一个列表为空时，函数将第一个列表的每个元素与第二个列表的每个元素进行比较，如果找到匹配项，它将加 1，然后除以大小以获得百分比。然而，当我运行这段代码时，我得到了很多错误：

main.hs:4:45: error:
• Couldn't match expected type ‘Float’ with actual type ‘Int’
• In the expression: (1 / length xs2) + fun2 tail xs1 xs2
  In the expression:
    if head xs1 == head xs2 then
        (1 / length xs2) + fun2 tail xs1 xs2
    else
        if head xs1 /= head xs2 then
            fun2 xs1 tail xs2
        else
            fun2 tail xs1 xs2
  In an equation for ‘fun2’:
      fun2 xs1 xs2
        = if head xs1 == head xs2 then
              (1 / length xs2) + fun2 tail xs1 xs2
          else
              if head xs1 /= head xs2 then
                  fun2 xs1 tail xs2
              else
                  fun2 tail xs1 xs2

main.hs:4:62: error:
• Couldn't match expected type ‘[Int] -> Int’
              with actual type ‘Float’
• The function ‘fun2’ is applied to three arguments,
  but its type ‘[Int] -> [Int] -> Float’ has only two
  In the second argument of ‘(+)’, namely ‘fun2 tail xs1 xs2’
  In the expression: (1 / length xs2) + fun2 tail xs1 xs2

main.hs:4:67: error:
• Couldn't match expected type ‘[Int]’
              with actual type ‘[a0] -> [a0]’
• Probable cause: ‘tail’ is applied to too few arguments
  In the first argument of ‘fun2’, namely ‘tail’
  In the second argument of ‘(+)’, namely ‘fun2 tail xs1 xs2’
  In the expression: (1 / length xs2) + fun2 tail xs1 xs2

main.hs:5:39: error:
• Couldn't match expected type ‘[Int] -> Float’
              with actual type ‘Float’
• The function ‘fun2’ is applied to three arguments,
  but its type ‘[Int] -> [Int] -> Float’ has only two
  In the expression: fun2 xs1 tail xs2
  In the expression:
    if head xs1 /= head xs2 then
        fun2 xs1 tail xs2
    else
        fun2 tail xs1 xs2

main.hs:5:48: error:
• Couldn't match expected type ‘[Int]’
              with actual type ‘[a1] -> [a1]’
• Probable cause: ‘tail’ is applied to too few arguments
  In the second argument of ‘fun2’, namely ‘tail’
  In the expression: fun2 xs1 tail xs2
  In the expression:
    if head xs1 /= head xs2 then
        fun2 xs1 tail xs2
    else
        fun2 tail xs1 xs2

main.hs:6:10: error:
• Couldn't match expected type ‘[Int] -> Float’
              with actual type ‘Float’
• The function ‘fun2’ is applied to three arguments,
  but its type ‘[Int] -> [Int] -> Float’ has only two
  In the expression: fun2 tail xs1 xs2
  In the expression:
    if head xs1 /= head xs2 then
        fun2 xs1 tail xs2
    else
        fun2 tail xs1 xs2

main.hs:6:15: error:
• Couldn't match expected type ‘[Int]’
              with actual type ‘[a2] -> [a2]’
• Probable cause: ‘tail’ is applied to too few arguments
  In the first argument of ‘fun2’, namely ‘tail’
  In the expression: fun2 tail xs1 xs2
  In the expression:
    if head xs1 /= head xs2 then
        fun2 xs1 tail xs2
    else
        fun2 tail xs1 xs2

main.hs:8:1: error:
• Couldn't match expected type ‘IO t0’ with actual type ‘Float’
• In the expression: main
  When checking the type of the IO action ‘main’

所以你能告诉我我做错了什么吗？

Answer 1

main.hs:4:45: error:
• Couldn't match expected type ‘Float’ with actual type ‘Int’
• In the expression: (1 / length xs2) + fun2 tail xs1 xs2
  …

length return 一个 Int.

例如，在 GHCi 中：

> :type length
length :: Foldable t => t a -> Int

> :set -XTypeApplications
> :type length @[]
length @[] :: [a] -> Int

（注意我写的是> 来表示提示。你可以使用:set prompt "> "和:set prompt-cont "| "来设置你的提示为相同的多行输入。）

虽然 (/) 适用于集合 Fractional 中的类型：

> :type (/)
(/) :: Fractional a => a -> a -> a

> :info Fractional
class Num a => Fractional a where
  (/) :: a -> a -> a
  recip :: a -> a
  fromRational :: Rational -> a
  {-# MINIMAL fromRational, (recip | (/)) #-}
    -- Defined in ‘GHC.Real’
instance forall a k (b :: k).
         Fractional a =>
         Fractional (Const a b)
  -- Defined in ‘Data.Functor.Const’
instance Fractional Float -- Defined in ‘GHC.Float’
instance Fractional Double -- Defined in ‘GHC.Float’

Int 不在 Fractional:

> :set -XFlexibleContexts
> () :: (Fractional Int) => ()

<interactive>:…:1: error:
    • No instance for (Fractional Int)
        arising from an expression type signature
    • In the expression: () :: (Fractional Int) => ()
      In an equation for ‘it’: it = () :: (Fractional Int) => ()

因此您需要将 length 的结果从 Int 转换为带有 fromIntegral 的 Float。此函数可以 return Num 中的任何类型，并注意上面 :info Fractional 的输出中 class Num a => Fractional a 意味着 Fractional 是 [= 的子集31=].

> :type fromIntegral
fromIntegral :: (Integral a, Num b) => a -> b

> fromIntegral (length "abc") :: Float
3.0

换句话说，你可以写成1 / fromIntegral (length xs2)。注意括号！这导致我们看到接下来的几条错误消息：

main.hs:4:62: error:
• Couldn't match expected type ‘[Int] -> Int’
              with actual type ‘Float’
• The function ‘fun2’ is applied to three arguments,
  but its type ‘[Int] -> [Int] -> Float’ has only two
  In the second argument of ‘(+)’, namely ‘fun2 tail xs1 xs2’
  …

当你写 fun2 tail xs1 xs2 时，这意味着“将 fun2 应用于三个参数：tail、xs1 和 xs2”。您希望将 tail xs1 的结果作为单个参数传递，因此您需要括号将它们组合在一起，即 fun2 (tail xs1) xs2。这也是下一个错误的原因：

main.hs:4:67: error:
• Couldn't match expected type ‘[Int]’
              with actual type ‘[a0] -> [a0]’
• Probable cause: ‘tail’ is applied to too few arguments
  In the first argument of ‘fun2’, namely ‘tail’
  In the second argument of ‘(+)’, namely ‘fun2 tail xs1 xs2’
  …

tail 是类型 [a] -> [a] 的函数，但您将其作为 fun2 的第一个参数传递，其第一个参数是类型 [Int] 的值.由于在 fun2.

的其他调用中出现相同的错误，相同的错误消息重复出现

例如，下一个表达式应为 fun2 xs1 (tail xs2)。下一个错误也有相同的根本原因，另外还为您提供了一个很好的解决方法提示：

main.hs:6:10: error:
• Couldn't match expected type ‘[Int] -> Float’
              with actual type ‘Float’
• <strong>The function ‘fun2’ is applied to three arguments,
  but its type ‘[Int] -> [Int] -> Float’ has only two</strong>
  In the expression: fun2 tail xs1 xs2
  …

最后，main 必须是一个 IO 动作，通常是 IO ()。它可能 return 任何类型的结果，也就是说，您可以对任何类型 t 使用 IO t，结果值将被简单地丢弃。但是，您当前传递的是 Float，调用 fun2 的结果，因此不匹配：

main.hs:8:1: error:
• Couldn't match expected type ‘IO t0’ with actual type ‘Float’
• In the expression: main
  When checking the type of the IO action ‘main’

解决办法是使用IO动作，例如print (fun2 [1,2,3,4] [3,4,5,6])，相当于putStrLn (show (fun2 [1,2,3,4] [3,4,5,6]))，两者都会将Float转换为String 使用调试转储 class Show 和 return 一个 IO 操作，它将在执行 main 时将结果打印到标准输出。

GHC 的错误消息并不总是完美的，但幸运的是，所有这些错误消息都包含足够的信息来解决您的问题。您只需要更多练习阅读它们并理解他们在说什么以及如何进行。

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