在Haskell中列出9个所有可能的4个选项 [英] List all possible 4 chooses from 9 in Haskell

问题描述

我无法找到一种从Haskell中的9个元素列表中挑选出4个元素的有效方法。
python-way做同样的事情：

  itertools.permutations（range（9 + 1），4 ）
  

在Haskell中不是很有效的方法：

  nub。 （地图（拿4））。 permutations $ [1..9] 
  

我想找到如下所示的内容：

  permutations 4 [1..9] 
  

<

  import Control.Arrow 
 
 select :: [a]  - > [（a，[a]）] 
选择[] = [] 
选择（x：xs）=（x，xs）：地图（第二个（x :)）（选择xs）
 
 perms :: Int  - > [a]  - > [[a]] 
 perm 0 _ = [[]] 
 perms n xs = do 
（y，ys）<  - 选择xs 
 fmap（y :)（ perm（n-1）ys）
  

它非常懒惰，甚至适用于无限列表，尽管输出没有什么用处。我没有打扰实现对角化或类似的东西。对于有限的列表，它很好。

I'm not able to find an effective way to pick out all permutations of 4 elements from a list of 9 elements in Haskell. The python-way to do the same thing:

itertools.permutations(range(9+1),4)

An not so effective way to do it in Haskell:

nub . (map (take 4)) . permutations $ [1..9]

I would like to find something like:

permutations 4 [1..9]

解决方案

Here is my solution:

import Control.Arrow

select :: [a] -> [(a, [a])]
select [] = []
select (x:xs) = (x, xs) : map (second (x:)) (select xs)

perms :: Int -> [a] -> [[a]]
perms 0 _  = [[]]
perms n xs = do
    (y, ys) <- select xs
    fmap (y:) (perms (n - 1) ys)

It's very lazy and even works for infinite lists, although the output there is not very useful. I didn't bother implementing diagonalization or something like that. For finite lists it's fine.

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