Haskell 函数式编程新手

嘿，我对Haskell真的是个新手，我的一生都在使用更经典的编程语言。我不知道这里发生了什么。我试图实现一个非常简单的维特比算法，但只针对两种状态（诚实和不诚实）

我有一个问题，我想解决我的数组，但我认为我没有得到正确的类型。或者我每次尝试寻址时都会创建一个新数组——同样愚蠢。请特别注意myArray、中缀和dynamicProgram。非常漂亮

Code

import Array
import Char

trans :: Int -> Int -> Double -> Double -> Double
trans from x trans11 trans21 =
    if (from == 1) && (x == 1)
        then trans11
    else if (from == 1) && (x == 2) 
        then (1-trans11)
    else if (from == 2) && (x == 1) 
        then trans21
    else (1-trans21)

em :: Char -> [Double] -> Double
em c list = list!! a
    where a = digitToInt c

intToChar :: Int -> Char
intToChar n | n == 1 = '1'
            | n == 2 = '2'

casino :: Char -> Int -> Int -> [Double] -> [Double] -> Double -> Double -> Double
casino seqchar 1 y em1 em2 t1 t2= 0.5 * (em seqchar em1)
casino seqchar 2 y em1 em2 t1 t2= 0.5 * (em seqchar em2)
casino seqchar x y em1 em2 t1 t2= maximum[ (1 @@ y-1)*(em seqchar em1)*(trans 1 x t1 t2),(2 @@ y-1)*(em seqchar em2)*(trans 2 x t1 t2) ]

dynamicProgram :: [Char] -> (Char -> Int -> Int -> [Double] -> [Double] -> Double -> Double -> Double) -> [Double] -> [Double] -> Double -> Double -> (Array a b)
dynamicProgram string score list1 list2 trans11 trans21 = myArray 1 len
                                [score (string!!y) x y list1 list2 trans11 trans21 | x  Int -> [Double] -> Array a b
myArray startIndex endIndex values = listArray (startIndex,startIndex) (endIndex,endIndex) values

traceback :: [Char] -> Int -> Int -> [Double] -> [Double] -> Double -> Double -> [Char]
traceback s 1 0 em1 em2 t1 t2 = []
traceback s 2 0 em1 em2 t1 t2 = []
traceback s x y em1 em2 t1 t2 | x@@y == (1 @@ y-1)*(em (s!!y) em1)*(trans 1 x t1 t2) = '1' : traceback s 1 (y-1) em1 em2 t1 t2
                            | x@@y == (2 @@ y-1)*(em (s!!y) em1)*(trans 2 x t1 t2) = '2' : traceback s 2 (y-1) em1 em2 t1 t2 

answer :: [Char] -> [Double] -> [Double] -> Double -> Double -> [Char]
answer string list1 list2 t1 t2 = reverse $ maxC : traceback string max end list1 list2 t1 t2 $ dynamicProgram casino string list1 list2 t1 t2
   where
      end = (length string) + 1
      max | maximum (1@@end) (2@@end) == 1@@end = 1
      | maximum (1@@end) (2@@end) == 2@@end = 2
      maxC = intToChar max

infix 5 @@
(@@) i j = myArray ! (i, j)

main = do
    putStrLn "What is the sequence to test?"
    seq  state 1 transmission probability?"
    trp1  state 2 transmission probability is " ++ (1-trp1)
    putStrLn "What is the state 2 -> state 1 transmission probability?"
    trp2  state 2 transmission probability is " ++ (1-trp2)
    putStrLn "I assume that the prob of starting in either state is 1/2.  Go!"
    answer seq st1 st2 trp1 trp2

---

我从编辑窗口复制了代码（stackoverflow解析器中的某些东西正在吞噬部分代码），并在ghci上进行了尝试，发现了几个错误。第一个错误是：

foo.hs:34:71:
    Couldn't match expected type `[e]' against inferred type `(a, b)'
    In the second argument of `listArray', namely
        `(endIndex, endIndex)'
    In the expression:
        listArray (startIndex, startIndex) (endIndex, endIndex) values
    In the definition of `myArray':
        myArray startIndex endIndex values
                  = listArray (startIndex, startIndex) (endIndex, endIndex) values

foo.hs:43:44:
    Couldn't match expected type `a -> b'
           against inferred type `[Char]'
    In the first argument of `($)', namely
        `maxC : (traceback string max end list1 list2 t1 t2)'
    In the second argument of `($)', namely
        `(maxC : (traceback string max end list1 list2 t1 t2))
       $ (dynamicProgram casino string list1 list2 t1 t2)'
    In the expression:
          reverse
        $ ((maxC : (traceback string max end list1 list2 t1 t2))
         $ (dynamicProgram casino string list1 list2 t1 t2))

foo.hs:51:11:
    Couldn't match expected type `Array i e'
           against inferred type `Int -> Int -> [Double] -> Array a b'
    In the first argument of `(!)', namely `myArray'
    In the expression: myArray ! (i, j)
    In the definition of `@@': @@ i j = myArray ! (i, j)

foo.hs:63:4:
    Couldn't match expected type `[a]' against inferred type `IO ()'
    In the first argument of `(++)', namely
        `putStrLn
           "I assume that the state 1 -> state 2 transmission probability is "'
    In the expression:
          (putStrLn
             "I assume that the state 1 -> state 2 transmission probability is ")
        ++
          (1 - trp1)
    In a 'do' expression:
          (putStrLn
             "I assume that the state 1 -> state 2 transmission probability is ")
        ++
          (1 - trp1)

listArray的类型为：

listArray :: (Ix i) => (i, i) -> [e] -> Array i e
        -- Defined in GHC.Arr

它需要一个具有上下界和列表的元组。因此，正确的表达式可能是：

listArray (startIndex, endIndex) values

myArray的类型不是

数组a b

，而是

数组Int Double

第二个错误是：

foo.hs:34:71:
    Couldn't match expected type `[e]' against inferred type `(a, b)'
    In the second argument of `listArray', namely
        `(endIndex, endIndex)'
    In the expression:
        listArray (startIndex, startIndex) (endIndex, endIndex) values
    In the definition of `myArray':
        myArray startIndex endIndex values
                  = listArray (startIndex, startIndex) (endIndex, endIndex) values

foo.hs:43:44:
    Couldn't match expected type `a -> b'
           against inferred type `[Char]'
    In the first argument of `($)', namely
        `maxC : (traceback string max end list1 list2 t1 t2)'
    In the second argument of `($)', namely
        `(maxC : (traceback string max end list1 list2 t1 t2))
       $ (dynamicProgram casino string list1 list2 t1 t2)'
    In the expression:
          reverse
        $ ((maxC : (traceback string max end list1 list2 t1 t2))
         $ (dynamicProgram casino string list1 list2 t1 t2))

foo.hs:51:11:
    Couldn't match expected type `Array i e'
           against inferred type `Int -> Int -> [Double] -> Array a b'
    In the first argument of `(!)', namely `myArray'
    In the expression: myArray ! (i, j)
    In the definition of `@@': @@ i j = myArray ! (i, j)

foo.hs:63:4:
    Couldn't match expected type `[a]' against inferred type `IO ()'
    In the first argument of `(++)', namely
        `putStrLn
           "I assume that the state 1 -> state 2 transmission probability is "'
    In the expression:
          (putStrLn
             "I assume that the state 1 -> state 2 transmission probability is ")
        ++
          (1 - trp1)
    In a 'do' expression:
          (putStrLn
             "I assume that the state 1 -> state 2 transmission probability is ")
        ++
          (1 - trp1)

$

是右关联的，因此首先查看最右边的

$

。它的第一个参数必须是一个函数，它将使用最右边的参数作为参数调用该函数。然而，这里是一个列表

第三个错误是：

foo.hs:34:71:
    Couldn't match expected type `[e]' against inferred type `(a, b)'
    In the second argument of `listArray', namely
        `(endIndex, endIndex)'
    In the expression:
        listArray (startIndex, startIndex) (endIndex, endIndex) values
    In the definition of `myArray':
        myArray startIndex endIndex values
                  = listArray (startIndex, startIndex) (endIndex, endIndex) values

foo.hs:43:44:
    Couldn't match expected type `a -> b'
           against inferred type `[Char]'
    In the first argument of `($)', namely
        `maxC : (traceback string max end list1 list2 t1 t2)'
    In the second argument of `($)', namely
        `(maxC : (traceback string max end list1 list2 t1 t2))
       $ (dynamicProgram casino string list1 list2 t1 t2)'
    In the expression:
          reverse
        $ ((maxC : (traceback string max end list1 list2 t1 t2))
         $ (dynamicProgram casino string list1 list2 t1 t2))

foo.hs:51:11:
    Couldn't match expected type `Array i e'
           against inferred type `Int -> Int -> [Double] -> Array a b'
    In the first argument of `(!)', namely `myArray'
    In the expression: myArray ! (i, j)
    In the definition of `@@': @@ i j = myArray ! (i, j)

foo.hs:63:4:
    Couldn't match expected type `[a]' against inferred type `IO ()'
    In the first argument of `(++)', namely
        `putStrLn
           "I assume that the state 1 -> state 2 transmission probability is "'
    In the expression:
          (putStrLn
             "I assume that the state 1 -> state 2 transmission probability is ")
        ++
          (1 - trp1)
    In a 'do' expression:
          (putStrLn
             "I assume that the state 1 -> state 2 transmission probability is ")
        ++
          (1 - trp1)

myArray

不是数组；它是一个函数，它接受三个参数并基于它们构造一个数组

在这里，您对更传统的命令式语言的习惯可能会让您感到困惑。在传统的命令式语言中，很自然会有一个全局

myArray

变量，您可以从程序的中间访问该变量。然而，在Haskell中，如果没有更高级的技巧（初学者不应该尝试），那么“全局”变量更像是一个常量值（在第一次使用时是惰性计算的，但在生成可执行文件时，编译器可能已经计算了它）。不能从作为输入读取的值初始化它

解决这个问题的最佳方法是将数组传递给程序，不幸的是，这个程序需要进行一些更改，并且否定了

@

操作符的有用性。您可以用几种更高级的方法隐藏数组的传递，但在学习过程中，最好是更加明确

最后一个错误是：

foo.hs:34:71:
    Couldn't match expected type `[e]' against inferred type `(a, b)'
    In the second argument of `listArray', namely
        `(endIndex, endIndex)'
    In the expression:
        listArray (startIndex, startIndex) (endIndex, endIndex) values
    In the definition of `myArray':
        myArray startIndex endIndex values
                  = listArray (startIndex, startIndex) (endIndex, endIndex) values

foo.hs:43:44:
    Couldn't match expected type `a -> b'
           against inferred type `[Char]'
    In the first argument of `($)', namely
        `maxC : (traceback string max end list1 list2 t1 t2)'
    In the second argument of `($)', namely
        `(maxC : (traceback string max end list1 list2 t1 t2))
       $ (dynamicProgram casino string list1 list2 t1 t2)'
    In the expression:
          reverse
        $ ((maxC : (traceback string max end list1 list2 t1 t2))
         $ (dynamicProgram casino string list1 list2 t1 t2))

foo.hs:51:11:
    Couldn't match expected type `Array i e'
           against inferred type `Int -> Int -> [Double] -> Array a b'
    In the first argument of `(!)', namely `myArray'
    In the expression: myArray ! (i, j)
    In the definition of `@@': @@ i j = myArray ! (i, j)

foo.hs:63:4:
    Couldn't match expected type `[a]' against inferred type `IO ()'
    In the first argument of `(++)', namely
        `putStrLn
           "I assume that the state 1 -> state 2 transmission probability is "'
    In the expression:
          (putStrLn
             "I assume that the state 1 -> state 2 transmission probability is ")
        ++
          (1 - trp1)
    In a 'do' expression:
          (putStrLn
             "I assume that the state 1 -> state 2 transmission probability is ")
        ++
          (1 - trp1)

这有两个错误：编译器抱怨的一个是优先级问题，正如编译器添加的括号清楚地显示的那样，通过正确应用括号或

$

运算符，可以很容易地解决这个问题。另一个错误是，在修复此错误后，您会发现，无法将字符串和数字连接起来；您必须将数字转换为字符串

这一切都没有考虑算法甚至大部分代码，只是考虑编译器错误。例如，如果需要二维数组，第一个错误的正确表达式为：

listArray ((startIndex, startIndex), (endIndex, endIndex)) values

---

现在，两个边界都是元组，其类型为

数组（Int，Int）Double

您可以像这样重写trans函数：

trans :: Int -> Int -> Double -> Double -> Double
trans 1 1 trans11 trans21 =   trans11
trans 1 2 trans11 trans21 = 1-trans11
trans 2 1 trans11 trans21 =   trans21
trans _ _ trans11 trans21 = 1-trans21