Check, if list is a sublist of another list

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有刺的猬
有刺的猬 2021-01-19 08:57

I\'d like to write a function that checks if one list is a sublist of another list. I wrote this, but it is not working, but I need something like this, I guess. Thanks for

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  • 2021-01-19 09:37

    You could use the function intersect like this:

    intersect [list1] [list2] == [list1]

    *Test> intersect [] [] == []
    True
    *Test> intersect [1] [] == [1]
    False
    *Test> intersect [2,4,6,4] [9,5,4,2,6,3,3] == [2,4,6,4]
    True
    *Test> intersect [1,2,3] [3,2,5,4] == [1,2,3]
    False
    
    

    You will have to import Data.List for this.

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  • 2021-01-19 09:58

    Your code is close to working, but just needs some minor changes. As other have said in the comments, you need to include | pattern guards, and remove = from the first function call. Here is what the last 3 lines should look like:

    subList (x:xs) (y:ys)   
        | x == y    = subList xs ys   
        | otherwise = subList (x:xs) ys
    

    This will fix up your code for the most part, but you also need to add the base case subList [] [] = True, because the empty list [] is the a sublist of another empty list [], just like [1] is a sublist of [1].

    Adding these changes, your code would look like this:

    subList :: Eq a => [a] -> [a] -> Bool
    subList [] [] = True
    subList _ []    = False
    subList [] _    = True
    subList (x:xs) (y:ys) 
        | x == y    = subList xs ys   
        | otherwise = subList (x:xs) ys
    

    Some example calls:

    Prelude> subList [] []
    True
    Prelude> subList [1] [1,2,3]
    True
    Prelude> subList [1] [4,2,3]
    False
    Prelude> subList [1] []
    False
    Prelude> subList [1,2] [1,2]
    True
    Prelude> subList [1,2] [2,1]
    False
    Prelude> subList [1,2] [1,2,2,1]
    True
    

    However, their is problem with a call like this:

    Prelude> subList [1,3] [1,2,3]
    True
    

    implying that [1,3] is a sublist of [1,2,3]. This could be intended, but if it is not, then you need to change your approach.

    Another approach:

    For your two lists, xs and ys, You could instead split the ys into sublists of length xs, let's say subys, and check if xs exists in subys. To do this, you could use splitAt, which splits a list every n characters. Here is an example function:

    split_lists :: Int -> [a] -> [[a]]
    split_lists _ [] = []
    split_lists n xs
        | length first == n = first : restxs
        | otherwise = restxs
        where (first, rest) = splitAt n xs
              restxs = split_lists n (tail first ++ rest)
    

    If you don't wish to use splitAt, you could do something like this:

    split_lists :: Int -> [a] -> [[a]]
    split_lists _ [] = []
    split_lists n xs = filter (\x -> length x == n) list
        where list = take n xs : split_lists n (drop 1 xs)
    

    Which behaves like:

    Prelude> split_lists 3 [1,2,3,4,5,6,7,8,9,10]
    [[1,2,3],[2,3,4],[3,4,5],[4,5,6],[5,6,7],[6,7,8],[7,8,9],[8,9,10]]
    

    Then you could just use any to check if the first list exists in the second list splitted into sublists, or you could just use normal recursion, up to you.

    Here is an example using any:

    subList :: (Eq a) => [a] -> [a] -> Bool
    subList [] [] = True
    subList xs ys = any (==xs) subys
        where subys = (split_lists (length xs) ys)
    

    Here is an example using recursion:

    subList :: (Eq a) => [a] -> [a] -> Bool
    subList [] [] = True
    subList xs ys = check_lists xs subys
        where subys = (split_lists (length xs) ys)
    
    check_lists :: (Eq a) => [a] -> [[a]] -> Bool
    check_lists _ [] = False
    check_lists xs (y:ys)
        | xs == y = True
        | otherwise = check_lists xs ys
    

    Which now behaves as follows:

    Prelude> subList [] []
    True
    Prelude> subList [1] [1,2,3]
    True
    Prelude> subList [1] [4,2,3]
    False
    Prelude> subList [1] []
    False
    Prelude> subList [1,2] [1,2]
    True
    Prelude> subList [1,2] [2,1]
    False
    Prelude> subList [1,2] [1,2,2,1]
    True
    Prelude> subList [1,3] [1,2,3]
    False
    Prelude> subList [1,2] [0,1,2,3]
    True
    
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