索引与[INT]和[字符串]]哈斯克尔

Question

我在指数[Int]和[[String]]，现在我只是想保持其对应指数[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？

我已经试过map (!!) (x y)

其中x是[字符串] y为[INT]

Answer 1

也许你正在寻找的东西是这样的：

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

类型也可以推广到

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

因为我们不需要列表o F-名单。

这不是非常有效。例如，如果我们可以假设指数正在增加，它可以得到很大的改善。

Answer 2

我想使用列表推导是这项工作的一种表达方式。

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"]] 

所以根据@ Centril的评论，我必须同意这里是一个单一的版本的同一件事;

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"]]