Question

我正在寻找一种技术，该技术允许在对之前的列表的后续折叠调用之间进行记忆。

我看过memoize library，但这似乎不支持对高阶函数的记忆，对于折叠来说就是这种情况。

我还尝试了使用懒惰的评估结果图的方法，但无济于事。

以下是简单的示例代码：

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

时间表明折痕是从头算起的。有没有办法使后续的折叠重用以前的折叠结果？

Answer 1

输入数据类型！

>>> float('NaN')
nan
>>> type(float('NaN'))
<class 'float'>

请注意，会导出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)类型，但不会导出List构造函数，因此构造List的唯一方法是使用List函数（通常称为“智能构造函数”。

记住后续折叠请求

1 个答案: