我如何在Haskell中分離元組？

Question

我如何合並一個元組列表而不重復那些元組中的任何項目？

例如 ：

從列表[（“ a”，“ b”），（“ c，” d“），（” a“，” b）]中，應返回[“ a”，“ b”，“ c”，“ d“]

---

所以我得到該代碼的錯誤消息：

No instance for (Eq a0) arising from a use of `nub'
The type variable `a0' is ambiguous
Possible cause: the monomorphism restriction applied to the following:
  merge :: [(a0, a0)] -> [a0] (bound at P.hs:9:1)
Probable fix: give these definition(s) an explicit type signature
              or use -XNoMonomorphismRestriction
Note: there are several potential instances:
  instance Eq a => Eq (GHC.Real.Ratio a) -- Defined in `GHC.Real'
  instance Eq () -- Defined in `GHC.Classes'
  instance (Eq a, Eq b) => Eq (a, b) -- Defined in `GHC.Classes'
  ...plus 22 others
In the first argument of `(.)', namely `nub'
In the expression: nub . mergeTuples
In an equation for `merge':
    merge
      = nub . mergeTuples
      where
          mergeTuples = foldr (\ (a, b) r -> a : b : r) []

失敗，模塊已加載：無。

Answer 1

讓我們將其分離出來，首先，合並元組

mergeTuples :: [(a, a)] -> [a]
mergeTuples = concatMap (\(a, b) -> [a, b]) -- Thanks Chuck
-- mergeTuples = foldr (\(a, b) r -> a : b : r) []

然后我們可以使用nub使其獨特

merge :: Eq a => [(a, a)] -> [a]
merge = nub . mergeTuples

如果你想所有人在一起

merge = nub . mergeTuples
  where mergeTuples = concatMap (\(a, b) -> [a, b])

或者，如果您想將其粉碎在一起（請勿這樣做）

merge [] = []
merge ((a, b) : r) = a : b : filter (\x -> x /= a && x /= b) (merge r)