后续折叠调用的记忆

Question

我正在寻找一种技术，该技术允许在针对前置列表的后续折叠调用之间进行记忆。

我查看了 memoize library，但这似乎不支持高阶函数的记忆，折叠就是这种情况。

我也尝试过使用惰性评估结果图的技术，但无济于事。

这是简单的示例代码：

module Main where

import Data.Time

printAndMeasureTime :: Show a => a -> IO ()
printAndMeasureTime a = do
  startTime <- getCurrentTime
  print a
  stopTime <- getCurrentTime
  putStrLn $ " in " ++ show (diffUTCTime stopTime startTime)

main = do
  let as = replicate 10000000 1
  printAndMeasureTime $ foldr (-) 0 as -- just to resolve thunks
  printAndMeasureTime $ sum as
  printAndMeasureTime $ sum (1:as) -- recomputed from scratch, could it reuse previous computation result?
  printAndMeasureTime $ length (as)
  printAndMeasureTime $ length (1:as) -- recomputed from scratch, could it reuse previous computation result?

和输出：

0
 in 1.125098223s
10000000
 in 0.096558168s
10000001
 in 0.104047058s
10000000
 in 0.037727126s
10000001
 in 0.041266456s

时代表明折叠是从头开始计算的。有没有办法让后续的折叠重复使用之前的折叠结果？

Answer 1

做一个数据类型！

module List (List, _elements, _sum, _length, toList, cons) where

data List = List
  { _elements :: [Int]
  , _sum :: !Int
  , _length :: !Int
  }

toList :: [Int] -> List
toList xs = List xs (sum xs) (length xs)

cons :: Int -> List -> List
cons x (List xs t n) = List (x:xs) (x+t) (1+n)

请注意 List 类型是导出的，但 List 构造函数不是，因此构造 List 的唯一方法是使用 toList 函数（通常称为 "smart constructor"）。

后续折叠调用的记忆

Memoization for subsequent fold calls

haskell

memoization

fold