haskell - 将列表拆分为可能的元组列表

Question

我需要将列表拆分为所有可能的元组列表，但我不确定如何执行此操作。

例如:

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

应该导致:
[("cat","dog"), ("cat","mouse"), ("dog","cat"), ("dog","mouse"), ("mouse","cat"), ("mouse","dog")]
我能够形成前两个，但我不确定如何获得其余的。

这是我到目前为止所拥有的:

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

Answer 1

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 ]