Question

我有索引[Int]和[[String]]，现在我只想保留与索引相对应的[String]：

[0,1,4]和[[a,a,a],[b,b,b],[c,c,c],[d,d,d],[e,e,e],[f,f,f]]

会给[[a,a,a],[b,b,b],[e,e,e]]

所以[Int] -> [[String]] -> [[String]]

我怎么能这样做？

我试过了map (!!) (x y)

其中x是[[String]]，y是[Int]

Answer 1

也许你正在寻找这样的东西：

foo :: [Int] -> [[String]] -> [[String]]
foo indices strings = map (strings !!) indices

该类型也可以推广到

foo :: [Int] -> [a] -> [a]

因为我们不需要列表清单。

这不是非常有效。例如，如果我们可以假设指数正在增加，那么可以大大改善。

Answer 2

我想使用列表推导是这项工作的一种富有表现力的方式。

getStrings :: [Int] -> [[String]] ->  [[String]]
getStrings is css = [cs | (ix,cs) <- zip [0..] css, ix `elem` is]

*Main> gs [0,1,4] [["a","a","a"],["b","b","b"],["c","c","c"],["d","d","d"],["e","e","e"],["f","f","f"]]
[["a","a","a"],["b","b","b"],["e","e","e"]]

根据@ Centril的评论，我必须同意这里是同一个版本的monadic版本;

gs :: [Int] -> [[String]] ->  [[String]]
gs is css = zip [0..] css >>= \(ix,cs) -> guard (ix `elem` is) >> return cs

*Main> gs [0,1,4] [["a","a","a"],["b","b","b"],["c","c","c"],["d","d","d"],["e","e","e"],["f","f","f"]]
[["a","a","a"],["b","b","b"],["e","e","e"]]

使用[Int]和[[String]] Haskell进行索引

2 个答案: