哈斯克尔列表理解寻找素数 [英] Haskell list comprehension for finding primes

问题描述

我试图使用列表推导尽可能简洁地查找小于某个整数 n 的所有素数。我正在学习Haskell，这只是一个练习。我想写一些东西：

  isqrt :: Integral a => a  - > 
 isqrt = floor。 sqrt。 fromIntegral 
 
 primes :: Integral a => a  - > [a] 
素数n = [i | i < -  [1,3..n]，mod i k / = 0 | k   

这当然不起作用。有没有一种方法可以在列表理解中获得列表理解？

这是我得到的错误：

 练习-99-1.hs：138：39：不在范围中：`k'
 
练习-99-1.hs：138：46 ：
非法平行列表理解：使用-XParallelListComp 
 
练习-99-1.hs：138：68：不在范围内：`i'
  
 
 
 
但是 - 我并不真正希望语法甚至是合法的： - ）
 
 
 
意图是尽可能直接翻译： 素数n  =奇数整数 i 小于 n 使得 i 不能被任何 k ，所有 k 在集合中： primes（isqrt i）  - 或多或少。 （我希望我得到的是对的？）
 
 
 
谢谢！     
 >我在以下方面取得了一些进展： 
 
 
 $ b 
  primes： ：积分a => a  - > [a] 
素数2 = [2] 
素数n = 2：[i |如果（mod ik / = 0）then True else False）
（primes（isqrt i））] 
  

有没有更简单的方法来编写lambda谓词？

编辑：是的，这要归功于评论中的评论！

  primes :: Integral a => a  - > [a] 
素数2 = [2] 
素数n = 2：[i | i < -  [3,5..n]，all（（/ = 0）.mod i）（primes（isqrt i））] 
  

I'm trying to find all the primes less than some integer n as concisely as possible, using list comprehensions. I'm learning Haskell, and this is just an exercise. I'd like to write something like:

isqrt :: Integral a => a -> a   
isqrt = floor . sqrt . fromIntegral

primes :: Integral a => a -> [a]  
primes n = [i | i <- [1,3..n], mod i k /= 0 | k <- primes (isqrt i)]

which of course doesn't work. Is there a way to have a list comprehension inside a list comprehension?

Here is the error I'm getting:

exercise-99-1.hs:138:39: Not in scope: `k'

exercise-99-1.hs:138:46:
    Illegal parallel list comprehension: use -XParallelListComp

exercise-99-1.hs:138:68: Not in scope: `i'

BUT - I wasn't really expecting the syntax to even be legit :-)

The intent was to translate as directly as possible: " primes n = the set of odd integers i less than n such that i is not divisible by any k, for all k in the set: primes (isqrt i)" - more or less. (I hope I got that right?)

Thanks!

解决方案

I made some progress with the following:

primes :: Integral a => a -> [a]  
primes 2 = [2]  
primes n = 2:[i | i <- [3,5..n], all (\k -> if (mod i k /= 0) then True else False) 
                                     (primes (isqrt i))]

Is there a shorter way to write the lambda predicate?

edit: Yes, there is, thanks to the remarks in the comments!

primes :: Integral a => a -> [a]  
primes 2 = [2]  
primes n = 2:[i | i <- [3,5..n], all ((/= 0) . mod i) (primes (isqrt i))]

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