Question

我想获得以下字符串的一些轮换列表＆＃34;我想今晚挣脱＃34;。限制是旋转不能以单词＆＃34; to＆＃34;开头。或者＆＃34;今晚＆＃34;。所以轮换列表是["I want to break free today", "want to break free tonight I", "break free tonight I want to", "free tonight I want to break"]。

我写了以下函数：

rotate :: [l] -> [l]
rotate [] = []
rotate (x:xs) = (x:xs) ++ head(x:xs)

rotate1 :: [a] -> [[a]]
rotate1 xs = take (length xs) (iterate rotate xs)

main = do
print $ rotate1(words("I want to break free tonight"))

运行此代码，我获得了所有可能的轮换，但是它们形成了一个列表，其列表中包含["want", "I", "to", "break", "free", "tonight"]等与字符串"want I to break free tonight"不同的元素。另外，我想看看如何删除以"to"，"tonight"开头的旋转。我尝试使用第二部分的过滤功能，但我没有设法解决问题。任何帮助/提示表示赞赏。我注意到我是Haskell的初学者。

Answer 1

运行此代码......

代码无法运行。它有类型错误。

首先，让我们修复格式化，以便更容易阅读，并删除额外的括号

rotate :: [l] -> [l]
rotate [] = []
rotate (x:xs) = (x:xs) ++ head (x:xs)

rotate1 :: [a] -> [[a]]
rotate1 xs = take (length xs) (iterate rotate xs)

main = print $ rotate1 (words "I want to break free tonight")

这很奇怪：

rotate (x:xs) = (x:xs) ++ head (x:xs)

首先，x:xs是整个列表，x是列表的头部。例如，rotate [1, 2, 3]变为：

rotate [1, 2, 3] = let x = 1
                       xs = [2, 3]
                   in (x:xs) ++ head (x:xs)

rotate [1, 2, 3] = (1:[2, 3]) ++ head (1:[2, 3])
rotate [1, 2, 3] = [1, 2, 3] ++ head [1, 2, 3]
rotate [1, 2, 3] = [1, 2, 3] ++ 1
                   -- type error

但++需要双方列表。你可能想要的是：

rotate (x:xs) = xs ++ [x]

这给了我们：

rotate [1, 2, 3] = let x = 1
                       xs = [2, 3]
                   in xs ++ [x]
rotate [1, 2, 3] = [2, 3] ++ [1]
rotate [1, 2, 3] = [2, 3, 1]

这与：

相同

rotate x = tail x ++ [head x]

对于你的其余问题......过滤器应该是直截了当的，因为有一个filter函数可以完全满足你的需要，而unwords函数会将单词列表转换回字符串。

Answer 2

您需要multiprocessing.Queue()中的intercalate :: [a] -> [[a]] -> [a]功能。

来自hackage docs：

  插入:: [a] - ＆gt; [[a]] - ＆gt; [α]

插入xs xss相当于（concat（intersperse xs xss））。它   将列表xs插入xss中的列表之间并连接   结果

在Data.List

ghci

Haskell中列表的旋转

2 个答案: