删除列表中的重复元素Haskell

Question

我的清单是[[String]]，看起来像[[“ A1”，“ A2”，“ A3”]，[“ A1”，“ A2”，“ B1”]，[“ A1”，“ A2” ，“ B2”] .....]

我的代码如下所示

List = [[x,y,z] | x <- l1, y<- l2, z <- l3]
            where l1 = ["A1","A2","A3","B1","B2","B3","C1","C2","C3","D1","D2","D3"];
                          l2 = ....;
                          l3 = ....

compareTo :: [String] -> [String] -> Bool
compareTo x y
            |length (intersect x y) == length x     =True
            |otherwise                              =False 

removeDuplication :: [[String]] -> [[String]]
removeDuplication (x:xs) = nubBy compareTo (x:xs)

在这种情况下，没有考虑元素的顺序，它们是重复的[“ A1”，“ A2”，“ A3”]和[“ A2”，“ A3”，“ A1”]。

我想使用'nubBy'和'compareTo'函数来构建我的removeDuplication函数，而且我不确定如何将一个元素与列表中的所有其他元素进行比较。

Answer 1

似乎您正在尝试生成列表，该列表是l1中所有String的排列。 因此，也许您不需要nub和compareTo。

list = [[x,y,z] | (x:xs) <- tails l1, (y:ys) <- tails xs, z <- ys]
  where l1 = ["A1","A2","A3","B1","B2","B3","C1","C2","C3","D1","D2","D3"]

Answer 2

您已经完成了所有零碎的工作来解决您的问题。

您只需要组成最后一行：

removeDuplication = nubBy compareTo

修正了其余的错字后，一切都应该正常工作。