Haskell:将列表拆分为两个新列表的元组 [英] Haskell: Splitting list into tuple of two new lists

问题描述

我很难弄清楚如何将一个Ints列表拆分为一个包含两个新列表的元组，以便每个元素(从第一个开始)进入第一个列表，而每个其他元素进入第二个列表.

I am having difficulty figuring out how to split a list of Ints into a tuple containing two new lists, such that every element (starting with first) goes into the first list and every other element in the second.

像这样:

split [] = ([],[])
split [1] = ([1],[])
split [1,2] = ([1],[2])
split [1,2,3] = ([1,3],[2])
split [1,2,3,4] = ([1,3],[2,4])

我试图递归地(使用警卫)并仅使用单个参数xs来完成此操作

I'm trying to accomplish this recursively(with guards) and only using the single argument xs

这是我不断收到错误消息的方法:

This is my approach that keeps getting error messages:

split :: [Int] -> ([Int],[Int])
split xs | length(xs) == 0 = ([],[])
         | length(xs) == 1 = (xs !! 0 : [],[])
         | length(xs) == 2 = (xs !! 0 : [], xs !! 1 : [])
         | otherwise = (fst ++ xs !! 0, snd ++ xs !! 1) ++ split(drop 2 xs))    

推荐答案

您的split函数返回一对，但在最后一种情况下，您对split的结果使用++.这将是类型错误，因为++在列表上起作用，而不是成对出现.还有一个类型错误，因为fst和snd是挑选一对元素的函数，但是使用它们是一种奇怪的方式.

Your split function returns a pair, but in the last case you are using ++ on the result of split. That will be a type error, since ++ works on lists, not pairs. There is also a type error because fst and snd are functions to pick out the elements of a pair, but you are using them is a strange way.

此外，使用模式匹配而不是长度.另外，不需要测试长度是否为2的情况，因为一般情况下会删除2个元素，从而使您降至空列表的基本情况.

Furthermore, use pattern matching instead of using length. Also, the case where you test if the length is 2 is not needed, since the general case removes 2 elements which takes you down to the base case of the empty list.

还可以通过使用类型变量a代替类型中的Int来使函数更通用.

You can also make your function more general by using a type variable a instead of Int in the type.

:添加了代码

split :: [a] -> ([a], [a])
split [] = ([], [])
split [x] = ([x], [])
split (x:y:xys) = (x:xs, y:ys) where (xs, ys) = split xys

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