Question

我的列表是[[String]]，它看起来像[[“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，其中measn [“A1”，“A2”，“A3”]和[“A2”，“A3”，“A1”]是重复。< / p>

我想使用'nubBy'和'compareTo'函数来构建我的removeDuplication函数，我不知道如何将元素与列表中的所有其他元素进行比较。

Answer 1

看起来您正在尝试生成列表，这些列表是l1中所有字符串的排列。所以也许你不需要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

你已经完成了所有的部分工作来解决你的问题。

你只需要撰写最后一行：

removeDuplication = nubBy compareTo

在修复剩余的拼写错误后，一切都应该正常。

删除列表Haskell中的重复元素

2 个答案: