问题描述
Haskell newb here
我正在解决haskell中的这个问题：
（**）消除列表元素的连续重复。
如果列表包含重复的元素，则应该用元素的单个副本替换它们。元素的顺序不应该改变。
示例：
*（compress'（aaaabccaadeeee））
（ABCADE）
解决方案（我必须查找）使用foldr：
compress'::公式a）=> [a] - > [a]
compress'xs = foldr（\x acc - > if x ==（head acc）then acc else x：acc）[last xs] xs
根据解决方案，此foldr采用两个参数x和acc。似乎所有的foldr都采用这些参数;这有什么例外吗？就像需要3个或更多的foldr？如果不是这样，那么这个惯例是否是冗余的，并且可以用更少的代码来编写公式？ / code>接受2个参数的函数，但这并不妨碍它接受3个参数的函数，前提是该函数具有正确的类型签名。
如果我们有一个函数
g :: x - > y - > z - > w
使用
foldr ::（a - > b - > b） - > b - > [a] - > b
我们想将 g 传递给 foldr ，则（a→b→b）→（x→y→z→w）（其中〜是类型相等）。由于 - > 是正确的关联，这意味着我们可以将 g 的签名写为
x - > y - > （z - > w）
和它的含义是一样的。现在我们已经产生了两个参数的函数，它们返回一个参数的函数。为了将它与 a - >类型统一起来， b - > b ，我们只需要排列参数：
a - > | x - >
b - > | y - >
b | （z - > w）
这意味着 b〜z - > ; w ，所以 y〜b〜z - > w 和 a〜x 所以 g 的类型必须是
g :: x - > （z→w）→> （z - > w）
暗示
foldr g ::（z - > w） - > [x] - > （z - > w）
这当然不是不可能的，但更不可能。我们的累加器是一个函数，对我来说，这要求用DiffLists来演示：
type DiffList a = [a] - > [a]
append :: a - > DiffList a - > DiffList a
append x dl = \ xs - > dl xs ++ [x]
reverse':: [a] - > [a]
reverse'xs = foldr append（const []）xs $ []
请注意， foldr append（const []）xs 返回一个我们应用于 [] 来反转列表的函数。在这种情况下，我们给了 [a] - >类型的函数的别名。 [a] 称为 DiffList ，但它与写下
append :: a - > （[a] - > [a]） - > [a] - > [a]
这是3个参数的函数。
Haskell newb here
I'm working on this problem in haskell:
(**) Eliminate consecutive duplicates of list elements.
If a list contains repeated elements they should be replaced with a single copy of the element. The order of the elements should not be changed.
Example:
* (compress '(a a a a b c c a a d e e e e))
(A B C A D E)
The solution (which I had to look up) uses foldr:
compress' :: (Eq a) => [a] -> [a]
compress' xs = foldr (\x acc -> if x == (head acc) then acc else x:acc) [last xs] xs
This foldr, according to the solution, takes two parameters, x and acc. It would seem like all foldr's take these parameters; is there any exception to this? Like a foldr that takes 3 or more? If not, is this convention redundant and can the formula be written with less code?
foldr takes a function of 2 arguments, but this doesn't prevent it from taking a function of 3 arguments provided that function has the right type signature.
If we had a function
g :: x -> y -> z -> w
With
foldr :: (a -> b -> b) -> b -> [a] -> b
Where we want to pass g to foldr, then (a -> b -> b) ~ (x -> y -> z -> w) (where ~ is type equality). Since -> is right associative, this means we can write g's signature as
x -> y -> (z -> w)
and its meaning is the same. Now we've produced a function of two parameters that returns a function of one parameter. In order to unify this with the type a -> b -> b, we just need to line up the arguments:
a -> | x ->
b -> | y ->
b | (z -> w)
This means that b ~ z -> w, so y ~ b ~ z -> w and a ~ x so g's type really has to be
g :: x -> (z -> w) -> (z -> w)
implying
foldr g :: (z -> w) -> [x] -> (z -> w)
This is certainly not impossible, although more unlikely. Our accumulator is a function instead, and to me this begs to be demonstrated with DiffLists:
type DiffList a = [a] -> [a]
append :: a -> DiffList a -> DiffList a
append x dl = \xs -> dl xs ++ [x]
reverse' :: [a] -> [a]
reverse' xs = foldr append (const []) xs $ []
Note that foldr append (const []) xs returns a function which we apply to [] to reverse a list. In this case we've given an alias to functions of the type [a] -> [a] called DiffList, but it's really no different than having written
append :: a -> ([a] -> [a]) -> [a] -> [a]
which is a function of 3 arguments.
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