选择列表的倒数第二个元素

Question

过了一段时间,我决定回去学习一些函数式编程.我这次决定选择Haskell,因为它的功能和语法.

目前我正在做一些练习而且我被困住了.我想编写一个函数,从列表中选择第二个最后一个元素,即给定[1,2,3,4]它将为3.

这是我的功能.

lastButOne xs 
    | null xs = xs 
    | length xs == 1 = xs 
    | length xs == 2 = lastButOne head xs
    | otherwise = lastButOne tail xs

不幸的是,它产生了一些错

Couldn't match expected type `[a]' with actual type `[a1] -> [a1]'
Relevant bindings include
  xs :: [a] (bound at lastButOne.hs:1:12)
  lastButOne :: [a] -> [a] (bound at lastButOne.hs:1:1)
Probable cause: `tail' is applied to too few arguments
In the first argument of `lastButOne', namely `tail'
In the expression: lastButOne tail xs

我尝试了一些parthness,比如(head xs)和(tail xs),但它没有帮助.

   Occurs check: cannot construct the infinite type: a ~ [a]
   Expected type: [[a]]
     Actual type: [a]
   Relevant bindings include
     xs :: [a] (bound at lastButOne.hs:1:12)
     lastButOne :: [a] -> [a] (bound at lastButOne.hs:1:1)
   In the first argument of `head', namely `xs'
   In the first argument of `lastButOne', namely `(head xs)'

后续行动:或者我应该写后续行动.好吧,我的原始想法是编写一个函数,如果列表长度为1,则生成head元素.所以考虑到Lee的解释,很容易想出以下内容:

lastPattern :: [a] -> Maybe a
lastPattern [] = Nothing
lastPattern [x] = Just x
lastPattern [x,_] = Just x
lastPattern (_:xs) = lastPattern xs

这是第一个问题.条件[x]和[x,_]的模式是什么？

我想做的下一件事就是用反向编写相同的函数(正如Paul Johnson所指出的那样).我很快想出了head (tail (reverse [1,2,3,4]))在REPL中似乎工作得很好的东西.但是当我开始进行一些编码时,我最终得到了

 lastRev :: [a] -> a
lastRev xs 
    | null xs = error "empty list"
    | length xs == 1 = error "too short"
    | otherwise = head (tail (reverse xs))

因为头部功能是head :: [a] -> a.上面的功能对我来说有点乱,所以说.有什么方法可以让它成为现实:: [a] -> Maybe a吗？这是第二个问题.

最后但并非最不重要 - 第三个问题.哪个功能在性能方面更好？我怎样才能在Haskell中测量它？

Answer 1

(head xs)返回一个a,你试图将它传递到lastButOne需要[a]参数的地方.head xs在这种情况下你可以直接返回.前两个案例也有问题,因为它们返回一个列表,而元素是必需的.由于在这种情况下没有这样的元素,您可能会返回错误:

lastButOne :: [a] -> a
lastButOne xs 
    | null xs = error "empty list"
    | length xs == 1 = error "list too short"
    | length xs == 2 = head xs
    | otherwise = lastButOne (tail xs)

更功能的解决方案是在函数类型中编码偏好并返回一个,Maybe a这样Nothing如果输入列表太短,你就可以返回:

lastButOne :: [a] -> Maybe a
lastButOne xs 
    | null xs = Nothing
    | length xs == 1 = Nothing
    | length xs == 2 = Just (head xs)
    | otherwise = lastButOne (tail xs)

最后,更好的解决方案是使用模式匹配而不是保护长度:

lastButOne :: [a] -> Maybe a
lastButOne [] = Nothing
lastButOne [_] = Nothing
lastButOne [x,_] = Just x
lastButOne (_:xs) = lastButOne xs