Question

我在Haskell中有一个列表列表。我希望从每个列表中获取一个元素时获得所有可能性。我目前所拥有的是

a = [ [1,2], [10,20,30], [-1,-2] ] -- as an example
whatIWant = [ [p,q,r] | p <- a!!0, q <- a!!1, r <- a!!2 ]

这就是我想要的。但是，这显然不是很好的代码，我正在寻找一种更好的编写列表理解的方法，这样就不会在代码中显示索引号（0,1,2）...这就是我和＃39;被卡住了。

我该怎么做？

Answer 1

使用一个函数（里面使用列表推导），我的解决方案是

combinations :: [[a]] -> [[a]]
combinations [] = []
combinations [l] = map (\ x -> [x]) l
combinations (x:xs) = combine (combinations [x]) (combinations xs)
    where combine a b = [ p ++ q | p <- a, q <- b ]

示例：

*Main> combinations [[1, 2, 3], [4, 5, 6]]
[[1,4],[1,5],[1,6],[2,4],[2,5],[2,6],[3,4],[3,5],[3,6]]
*Main> combinations [['a', 'b', 'c'], ['A', 'B', 'C'], ['1', '2']]
["aA1","aA2","aB1","aB2","aC1","aC2","bA1","bA2","bB1",...
 "bB2","bC1","bC2","cA1","cA2","cB1","cB2","cC1","cC2"]

编辑：当然，您可以使用sequence功能，如评论中所示：

*Main> sequence [['a', 'b', 'c'], ['A', 'B', 'C'], ['1', '2']]
["aA1","aA2","aB1","aB2","aC1","aC2","bA1","bA2","bB1",...
 "bB2","bC1","bC2","cA1","cA2","cB1","cB2","cC1","cC2"]

Answer 2

这显然不是一个好的代码

这是关于你可以做到的最好的方法，因为你的约束是输入是一个列表列表。

如果您使用其他类型，例如三个列表，然后您可以在结构上索引。 E.g。

Prelude> let x@(a,b,c) = ( [1,2],  [10,20,30], [-1,-2] )

让你写：

Prelude> [ (p,q,r) | p <- a , q <- b , r <- c ]
[(1,10,-1),(1,10,-2),(1,20,-1)
,(1,20,-2),(1,30,-1),(1,30,-2)
,(2,10,-1),(2,10,-2),(2,20,-1)
,(2,20,-2),(2,30,-1),(2,30,-2)]

课程：要避免编制索引，请使用其结构捕获您想要保留的不变量的类型。将数据的维度提升到其类型。

清单列表理解从列表清单中抽样？

2 个答案: