扫描仪如何工作？哈斯克尔

Question

我一直在搞乱一些 Haskell 函数，有些我明白，有些不明白。

例如，如果我们这样做：scanl (+) 0 [1..3] 我的理解如下：

1. the accumulator is 0                  acc         = 0    |
2. (+) applied to acc and first el       acc = 0 + 1 = 1    |
3. (+) applied to latest acc and snd el  acc = 1 + 2 = 3    |
4. (+) applied to latest acc and third   acc = 3 + 3 = 6    V

现在，当我们列出列表时，我们会得到[0, 1, 3, 6]。

但我似乎无法理解scanr (+) 0 [1..3] 是如何给我的：[6,5,3,0] 也许scanr 的工作方式如下？

1. the first element in the list is the sum of all other + acc
2. the second element is the sum from right to left (<-) of the last 2 elements
3. the third element is the sum of first 2...

我不知道这是否是模式。

Answer 1

scanr 对应于foldr，scanl 对应于foldl。 foldr 从右边工作：

foldr (+) 0 [1,2,3] =
  (1 + (2 + (3 +   0))) =
  (1 + (2 +    3)) =
  (1 +    5) =
     6
-- [ 6,   5,   3,   0 ]

而scanr 只是按顺序显示中间结果：[6,5,3,0]。可以定义为

scanr (+) z xs = foldr g [z] xs
  where
  g x ys@(y:_) = x+y : ys

scanl 虽然应该像这样工作

scanl (+) 0 [1,2,3] =
  0 : scanl (+) (0+1) [2,3] =
  0 : 1 : scanl (+) (1+2) [3] =
  0 : 1 : 3 : scanl (+) (3+3) [] =
  0 : 1 : 3 : [6]

应该是这样的

scanl (+) z xs = foldr f h xs z
   where h      z = [z]
         f x ys z = z : ys (z + x)

Answer 2

scanl 和scanr 用于显示每次迭代时累加器的值。 scanl 从左到右迭代，scanr 从右到左迭代。

考虑以下示例：

scanl (+) 0 [1, 2, 3]

-- 0. `scanl` stores 0 as the accumulator and in the output list [0]
-- 1. `scanl` adds 0 and 1 and stores 1 as the accumulator and in the output list [0, 1]
-- 2. `scanl` adds 1 and 2 and stores 3 as the accumulator and in the output list [0, 1, 3]
-- 3. `scanl` adds 3 and 3 and stores 6 as the accumulator and in the output list [0, 1, 3, 6]
-- 4. `scanl` returns the output list [0, 1, 3, 6]

如您所见，scanl 在迭代列表时存储累加器的结果。 scanr 也是如此，但列表是反向迭代的。

这是另一个例子：

scanl (flip (:)) [] [1, 2, 3]

-- [[], [1], [2,1], [3,2,1]]

scanr       (:)  [] [1, 2, 3]

-- [[1,2,3], [2,3], [3], []]