是否有内置函数来获取 Haskell 中列表大小为 n 的所有连续子序列? [英] Is there a built-in function to get all consecutive subsequences of size n of a list in Haskell?

问题描述

例如，我需要一个函数:

For example, I need a function:

gather :: Int -> [a] -> [[a]]
gather n list = ???

where gather 3 "Hello!"== ["Hel","ell","llo","ol!"].

我有一个有效的实现:

gather :: Int-> [a] -> [[a]]
gather n list = 
    unfoldr 
        (x -> 
            if fst x + n > length (snd x) then 
                Nothing 
            else 
                Just 
                    (take 
                        n 
                        (drop 
                            (fst x)
                            (snd x)), 
                    (fst x + 1, snd x))) 
        (0, list)

但我想知道语言中是否已经为此内置了一些东西?我扫描了 Data.List 但没有看到任何东西.

but I am wondering if there is something already built into the language for this? I scanned Data.List but didn't see anything.

推荐答案

你可以使用tails:

gather n l = filter ((== n) . length) $ map (take n) $ tails l

或使用 takeWhile 而不是 filter:

gather n l = takeWhile ((== n) . length) $ map (take n) $ tails l

您可以通过删除从 tails 返回的列表的最后一个 n 元素来删除过滤步骤，如评论中所建议:

You can remove the filter step by dropping the last n elements of the list returned from tails as suggested in the comments:

gather n = map (take n) . dropLast n . tails
  where dropLast n xs = zipWith const xs (drop n xs)

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