来自列表性能的长度为 n 的子序列

问题描述

我实现了这个答案的一个版本https://stackoverflow.com/a/9920425/1261166(我不不知道回答者的意图是什么)

I implemented a version of this answer https://stackoverflow.com/a/9920425/1261166 (I don't know what was intended by the person answering)

sublistofsize 0 _        = [[]]
sublistofsize _ []       = []
sublistofsize n (x : xs) = sublistsThatStartWithX ++ sublistsThatDontStartWithX
  where sublistsThatStartWithX = map (x:) $ sublistofsize (n-1) xs
        sublistsThatDontStartWithX = sublistofsize n xs

我不确定的是 sublistsThatStartWithX = map (x:) $ sublistofsize (n-1) xs

我假设 map (x:) 给出了一个明智的性能问题，但不确定如何解决它.我已经对 print $ length $ sublistofsize 5 $ primesToTakeFrom 50

I assume that map (x:) gives a problem performance wise, but not sure of how to solve it. I've done profiling on print $ length $ sublistofsize 5 $ primesToTakeFrom 50

COST CENTRE                                  MODULE                                        no.     entries  %time %alloc   %time %alloc
sublistofsize                             Main                                          112     4739871   46.9   39.9    96.9  100.0
 sublistofsize.sublistsThatDontStartWithX Main                                          124     2369935    2.2    0.0     2.2    0.0
 sublistofsize.sublistsThatStartWithX     Main                                          116     2369935   47.8   60.1    47.8   60.1

我是否以一种好的方式实施了它?有什么更快的方法吗?

Did I implement it in a good way?Are there any faster ways of doing it?

推荐答案

没有.map 被高效编码并在线性时间内运行，这里没有问题.

No. map is coded efficiently and runs in linear time, no problems here.

但是，您的递归可能是一个问题.你们都在调用 sublistofsize (n-1) xs 和 sublistofsize n xs，它们 - 给定一个起始列表 sublistofsize m (_:_:ys) - 确实对术语 sublistofsize (m-1) ys 求值两次，因为它们在不同的递归步骤中没有共享.

However, your recursion might be a problem. You're both calling sublistofsize (n-1) xs and sublistofsize n xs, which - given a start list sublistofsize m (_:_:ys) - does evaluate the term sublistofsize (m-1) ys twice, as there is no sharing between them in the different recursive steps.

所以我会应用动态规划来获得

So I'd apply dynamic programming to get

subsequencesOfSize :: Int -> [a] -> [[a]]
subsequencesOfSize n xs = let l = length xs
                          in if n>l then [] else subsequencesBySize xs !! (l-n)
 where
   subsequencesBySize [] = [[[]]]
   subsequencesBySize (x:xs) = let next = subsequencesBySize xs
                             in zipWith (++) ([]:next) (map (map (x:)) next ++ [[]])

并不是说附加空列表是最漂亮的解决方案，但是您可以看到我如何将 zipWith 与置换列表一起使用，以便使用 next 的结果两次 - 一次直接在长度为 n 的子序列列表中，一次在长度为 n+1 的子序列列表中.

Not that appending the empty lists is the most beautiful solution, but you can see how I have used zipWith with the displaced lists so that the results from next are used twice - once directly in the list of subsequences of length n and once in the list of subsequences of length n+1.

在 GHCI 中使用 :set +s 对其进行测试，您会发现这比简单的解决方案要快得多:

Testing it in GHCI with :set +s, you can see how this is drastically faster than the naive solutions:

*Main> length $ subsequencesOfSize 7 [1..25]
480700
(0.25 secs, 74132648 bytes)
(0.28 secs, 73524928 bytes)
(0.30 secs, 73529004 bytes)
*Main> length $ sublistofsize 7 [1..25] -- @Vixen (question)
480700
(3.03 secs, 470779436 bytes)
(3.35 secs, 470602932 bytes)
(3.14 secs, 470747656 bytes)
*Main> length $ sublistofsize' 7 [1..25] -- @Ganesh
480700
(2.00 secs, 193610388 bytes)
(2.00 secs, 193681472 bytes)
*Main> length $ subseq 7 [1..25] -- @user5402
480700
(3.07 secs, 485941092 bytes)
(3.07 secs, 486279608 bytes)

这篇关于来自列表性能的长度为 n 的子序列的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！