Haskell如何从两个INT列表中返回INT列表

发布于 2025-02-10 04:39:07 字数 528 浏览 1 评论 0原文

如果list1 = [14,24,1,2,11,7,23,8,12,22,22,22,20,15,19,19,4,9,10,18,18,17,3,3,13 ，16,5,6,25]，和list2 = [14,14,24,24,1,1,2,2] 我如何返回的int列表[0,0,1,1,1,2,2,3,3]

我已经使用了该函数

elemIndex' :: Eq a => a -> [a] -> Int
elemIndex' x = fromMaybe (-1) . elemIndex x

，并且为第一个函数提供了结果 0 从列表1获取单位数字的索引但是我想要的是

findIndex :: [Int] -> [Int] -> [Int]

将输入两个int列表并返回键列表中的索引值基础我如何递归获取一个列表的索引或使用任何导入功能，例如MAP。

原文

if list1 = [14,24,1,2,11,7,23,8,12,22,20,0,15,19,4,9,10,21,18,17,3,13,16,5,6,25],
and list2 = [14,14,24,24,1,1,2,2]
How could I return a int list which is [0,0,1,1,2,2,3,3]

I have used the function

elemIndex' :: Eq a => a -> [a] -> Int
elemIndex' x = fromMaybe (-1) . elemIndex x

and the result for this for the first one
0
to get the index from list 1 for a single digit
But what I want is

findIndex :: [Int] -> [Int] -> [Int]

That would input two int list and return the index value base on the key list
How could I recursively get the index for one list or using any import function, such as map.

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

稚气少女 2025-02-17 04:39:07

您在解决问题方面做得很好。您已经认识到，您对第二个列表的每个元素的每个元素都做同样的事情，并且已经写了一个函数来捕获该行为。

elemIndex' :: Eq a => a -> [a] -> Int
elemIndex' x = fromMaybe (-1) . elemIndex x

现在，您需要获取此功能，并将其应用于另一个列表的每个元素，将结果收集到新列表中。也就是说，您想要看起来像的东西

mystery :: (a -> b) -> [a] -> [b]

，我们可以搜索 hoogle ，并确切找到该功能：

map :: (a -> b) -> [a] -> [b]

>代码>地图获取功能和列表，并将其应用于每个元素。您有一个函数（Elemindex'，带有适当的部分应用程序），并且您的示例中有一个列表（list2）。因此，让我们把它放在一起。

findAll :: Eq a => [a] -> [a] -> [Int]
findAll haystack needles = map (\x -> elemIndex' x haystack) needles

Haskell中有很多流函数，例如MAP < / code>，并且知道 /弄清楚在给定情况下使用哪种功能。如果您有疑问，请记住，您可以使用Hoogle来搜索一种类型，也可以简单地做任何要递归做的事情（如果您不知道MAP，则可以自己写一些递归），随着时间的流逝，您会掌握它。祝您在Haskell努力中祝您好运！

You've done a very good job of breaking the problem down. You've already recognized that you're doing the same thing for each element of the second list, and you've written a function to capture that behavior.

elemIndex' :: Eq a => a -> [a] -> Int
elemIndex' x = fromMaybe (-1) . elemIndex x

Now you want to take this function and apply it to each element of another list, collecting the results into a new list. That is, you want something that looks like

mystery :: (a -> b) -> [a] -> [b]

And we can search Hoogle and find exactly that function: map

map :: (a -> b) -> [a] -> [b]

map takes a function and a list and applies it to each element. You've got a function (elemIndex', with appropriate partial application), and you've got a list (list2 in your example). So let's put it together.

findAll :: Eq a => [a] -> [a] -> [Int]
findAll haystack needles = map (\x -> elemIndex' x haystack) needles

There are lots of streaming functions in Haskell like map, and knowing / figuring out which one to use in a given situation comes with practice. If you're ever in doubt, remember that you can use Hoogle to search for a type or you can simply do whatever you're trying to do recursively (if you didn't know map existed, you could write it yourself with a bit of recursion), and you'll get the hang of it over time. Good luck in your Haskell endeavors!

回复收藏 0 原文

岁月无声 2025-02-17 04:39:07

作为@silviomayolo答案的替代方法，您可以使用MAP数据结构。我们在list1中的每个值的索引中zip，然后将其转换为映射，然后在list2中对每个值进行查找。

import Data.Map
import Data.List

list1 = [14,24,1,2,11,7,23,8,12,22,20,0,15,19,4,9,10,21,18,17,3,13,16,5,6,25]
list2 = [14,14,24,24,1,1,2,2]

map1 = Data.Map.fromList $ list1 `zip` [0..]
-- fromList [(0,11),(1,2),(2,3),(3,20),(4,14),(5,23),(6,24),
--           (7,5),(8,7),(9,15),(10,16),(11,4),(12,8),(13,21),
--           (14,0),(15,12),(16,22),(17,19),(18,18),(19,13),
--           (20,10),(21,17),(22,9),(23,6),(24,1),(25,25)]

list3 = Data.List.map (map1 !) list2
-- [0,0,1,1,2,2,3,3]

As an alternative to the answer by @SilvioMayolo, you could use a Map data structure. We zip in the indexes for each value in list1 and then turn that into a map, then just do a lookup for each value in list2.

import Data.Map
import Data.List

list1 = [14,24,1,2,11,7,23,8,12,22,20,0,15,19,4,9,10,21,18,17,3,13,16,5,6,25]
list2 = [14,14,24,24,1,1,2,2]

map1 = Data.Map.fromList $ list1 `zip` [0..]
-- fromList [(0,11),(1,2),(2,3),(3,20),(4,14),(5,23),(6,24),
--           (7,5),(8,7),(9,15),(10,16),(11,4),(12,8),(13,21),
--           (14,0),(15,12),(16,22),(17,19),(18,18),(19,13),
--           (20,10),(21,17),(22,9),(23,6),(24,1),(25,25)]

list3 = Data.List.map (map1 !) list2
-- [0,0,1,1,2,2,3,3]

回复收藏 0 原文

~没有更多了~