如何在Haskell中获取无限列表中的每个第N个元素？ [英] How to get every Nth element of an infinite list in Haskell?

问题描述

更具体地说，我该如何从现有的无限列表中生成每个第N个元素的新列表？ 如果列表是[5,3,0,1,8,0,3,4,0,93,211,0 ...]，则获取每个第三元素将导致该列表[0,0,0,0 ，0 ...]

解决方案

我的版本使用drop：

 每n xs =案例下降（n-1）xs 
（y：ys） - > y：every n ys 
 []  - > [] 
  

编辑：这也适用于有限列表。只有无限列表，查尔斯斯图尔特的解决方案略短。

More specifically, how do I generate a new list of every Nth element from an existing infinite list?

E.g. if the list is [5, 3, 0, 1, 8, 0, 3, 4, 0, 93, 211, 0 ...] then getting every 3rd element would result in this list [0,0,0,0,0 ...]

解决方案

My version using drop:

every n xs = case drop (n-1) xs of
              (y:ys) -> y : every n ys
              [] -> []

Edit: this also works for finite lists. For infinite lists only, Charles Stewart's solution is slightly shorter.

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