列表使用列表理解的子列表 [英] Sublists of a list using list comprehension

问题描述

那很简单。我想使用list comprehension生成一个列表的所有子列表。

即：getSublist [1,2,3]是[[1]，[2]，[3 ]，[1,2]，[1,3]，[2,3]，[1,2,3]]

谢谢

解决方案

这已经被实现为 Data.List.subsequences ，但是如果你想要定义它你可以这样做：

你不能仅仅使用列表推导来完成它，但是通过一些递归，它看起来像这样： / p>

  sublists [] = [[]] 
 sublists（x：xs）= [x：sublist |子列表<  - 子列表xs] ++子列表xs 
  

阅读：空列表的唯一子列表是空的列表。 x：xs 的子列表（即头部 x 和尾部 xs的列表）的所有子列表都是 xs 以及每个 xs x >。

That simple. I want to generate all sublists of a list using list comprehension.

i.e: getSublist [1,2,3] is [[1], [2], [3], [1,2], [1,3], [2, 3], [1,2,3]]

Thanks

解决方案

This is already implemented as Data.List.subsequences, but if you want to define it yourself (for learning purposes), you can do it like this:

You can't do it with only list comprehensions, but with some recursion it looks like this:

sublists [] = [[]]
sublists (x:xs) = [x:sublist | sublist <- sublists xs] ++ sublists xs

Read: The only sublist of the empty list is the empty list. The sublists of x:xs (i.e. the list with the head x and the tail xs) are all of the sublists of xs as well as each of the sublists of xs with x prepended to them.

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