在Haskell中获取一个子列表 [英] Get a sublist in Haskell

问题描述

可能很简单，但我浏览过文档并以Google搜索为例，我仍然不确定答案。

如果我有像这样的列表：

  [1,2,3,4,5,6,7,8,9,0] 
  

我想提取一个切片，从索引4到索引8，即我想要的：

  [5,6,7,8,9] 
  

在Haskell中做到这一点的惯用方法是什么？

首先，这不是一个数组，它是一个列表。我不是（仅）迂腐，因为数组在Haskell中的问题比列表中多得多。

也就是说，一种常见的方法是使用 和 drop 在一起：



 前奏>下降4。取9美元[1,2,3,4,5,6,7,8,9,0] 
 [5,6,7,8,9] 
前奏>拿（9-4）。 drop 4 $ [1,2,3,4,5,6,7,8,9,0] 
 [5,6,7,8,9] 
  

后者效率更高一些。

Probably an easy one, but I've looked through the docs and googled for examples and I'm still not sure of the answer.

If I have a list like this:

[1,2,3,4,5,6,7,8,9,0]

and I want to extract a slice, say from index 4 to index 8 i.e. I want:

[5,6,7,8,9]

What is the idiomatic way to do this in Haskell?

解决方案

First of all, that's not an array, it's a list. I'm not being (merely) pedantic, as arrays are much more problematic in Haskell than lists.

That said, one common way is to use take and drop together:

Prelude> drop 4 . take 9 $ [1,2,3,4,5,6,7,8,9,0]
[5,6,7,8,9]
Prelude> take (9-4) . drop 4 $ [1,2,3,4,5,6,7,8,9,0]
[5,6,7,8,9]

The latter is a bit more efficient.

这篇关于在Haskell中获取一个子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！