list - 在 Haskell 中完全消除重复项

Question

我有这个代码，但它完全没有做我想要的，我需要一个元组列表；

[(3,2),(1,2),(1,3),(1,2),(4,3),(3,2),(1,2)]

并给

[(1,3),(4,3),(3,2),(1,2)]

但我希望它给予

[(1,3),(4,3)]

我哪里做错了？提前致谢。

eliminate :: [(Int,Int)] -> [(Int,Int)]
eliminate [] = []
eliminate (x:xs)
    | isTheSame xs x  = eliminate xs
    | otherwise       = x : eliminate xs


isTheSame :: [(Int,Int)] -> (Int,Int) -> Bool
isTheSame [] _ = False
isTheSame (x:xs) a
    | (fst x) == (fst a) && (snd x) == (snd a)  = True
    | otherwise                 = isTheSame xs a

Answer 1

代码几乎是正确的。只需改变这一行

    | isTheSame xs x  = eliminate xs

到

    | isTheSame xs x  = eliminate $ filter (/=x) xs   

原因是如果 x包含在 xs ，要删除所有出现的 x .

也就是说，您的代码示例中有几个部分可以更优雅地表达:

(fst x) == (fst a) && (snd x) == (snd a)与 x == a 相同

isTheSame与 elem 相同, 只是它的参数颠倒了

因此，我们可以表达函数 eliminate像这样:

eliminate [] = []
eliminate (x:xs)
  | x `elem` xs = eliminate $ filter (/=x) xs
  | otherwise = x : eliminate xs