将列表拆分成可能的元组列表 [英] Splitting list into a list of possible tuples

问题描述

我需要将列表分成所有可能的元组列表，但我不确定如何去做。

例如

 对[cat，dog，mouse] 
  

会导致

[（cat，dog），（cat ，鼠标），（狗，猫），（狗，鼠标），（鼠标，猫），（鼠标，狗）]



我能够形成前两个，但是我不确定如何获得剩下的。

  pairs :: [a]  - > [（a，a）] 
对（x：xs）= [（m，n）| m < -  [x]，n < -  xs] 
  

解决方案

您可以使用列表理解：

  allpairs :: Eq a => [a]  - > [（a，a）] 
 allpairs xs = [（x1，x2）| x1 < -  xs，x2 < -  xs，x1 / = x2] 
  

I need to split a list into a list of all possible tuples, but I'm unsure of how to do so.

For example

pairs ["cat","dog","mouse"]

should result in

[("cat","dog"), ("cat","mouse"), ("dog","cat"), ("dog","mouse"), ("mouse","cat"), ("mouse","dog")]

I was able to form the first two, but am unsure of how to get the rest.

Here's what I have so far:

pairs :: [a] -> [(a,a)]
pairs (x:xs) = [(m,n) | m <- [x], n <- xs]

解决方案

You can use a list comprehension:

allpairs :: Eq a => [a] -> [(a,a)]
allpairs xs = [ (x1,x2) | x1 <- xs, x2 <- xs, x1 /= x2 ]

这篇关于将列表拆分成可能的元组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！