在Haskell中过滤无限列表 [英] Filtering an infinite list in Haskell

问题描述

我有以下代码实现了Eratosthenes的筛选：

$ $ p $ primes :: [Int]
素数= primes'[2 ..]

primes':: [Int] - > [p] p'（p'ps），而不是（p'`isMultiple` p [prime]'[p] ）]）

a`isMultiple` b =（a`mod` b）== 0

main = print（sum（primes'[2..100000]））


我想将main改为

  main = print（sum [p | p   

毫不奇怪，这是因为它必须将p与无限列表素数的每个元素进行比较。既然我知道质数的顺序是递增的，当我发现一个超过我的上限的元素时，我该如何截断无限列表？

p.s。理论上，素数会过滤输入列表以返回素数列表。我知道如果我从2以外的东西开始列表，我将会遇到一些问题。我仍在努力如何自己解决这个问题，所以请不要破坏。谢谢; - ）

解决方案

在像这样的情况下，您知道一旦谓词对元素返回false，对于后面的元素会返回true，您可以用 takeWhile 来替换 filter ，它会在谓词停止时立即获取元素第一次返回false。

I have the following code that implements the Sieve of Eratosthenes:

primes :: [Int]
primes = primes' [2..]

primes' :: [Int] -> [Int]
primes' [] = []
primes' (p:ps) = p:(primes' [p' | p' <- ps, not (p' `isMultiple` p)])

a `isMultiple` b = (a `mod` b) == 0

main = print (sum (primes' [2..100000]))

I would like to change main to something like

main = print (sum [p | p <- primes, p < 100000]))

Not surprisingly, this hangs because it must compare p against every element of the infinite list primes. Since I know that primes is in increasing order, how do I truncate the infinite list as soon as I find an element that exceeds my upper limit?

p.s. In theory, primes' filters the input list to return a list of primes. I know there will be some issues if I start the list at something other than 2. I'm still working on how to address this issue on my own, so please no spoilers. Thanks ;-)

解决方案

In cases like this where you know that once the predicate returns false for an element, it won't ever return true for a later element, you can replace filter with takeWhile, which stops taking elements as soon as the predicate returns false for the first time.

这篇关于在Haskell中过滤无限列表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IT屋！