按索引交换列表中的两个元素 [英] Swap two elements in a list by its indices

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问题描述

有什么方法可以交换列表中的两个元素,如果我知道关于这些元素的唯一信息就是它们出现在列表中的位置。



要更具体一点,我正在寻找这样的东西:

  swapElementsAt :: Int  - > Int  - > [Int]  - > [b] 

表现如下:

 > swapElementsAt 1 3 [5,4,3,2,1]  - 交换第一个和第三个元素
[3,4,5,2,1]

我认为内置函数可能存在于Haskell中,但我无法找到它。

Haskell没有这样的功能,主要是因为它有点不起作用。你真的想要实现什么?



你可以实现你自己的版本(也许有更习惯写法的方法)。请注意,我假设 i< j ,但是扩展函数以正确处理其他情况将是微不足道的:

  swapElementsAt :: Int  - > Int  - > [a]  - > [a] 
swapElementsAt i j xs = let elemI = xs !!我
elemJ = xs !! (j + 1)xs)
right = drop(j + 1)xs
in left ++ [elemJ] ++ middle ++ [elemI] ++ right


Is there any way to swap two elements in a list if the only thing I know about the elements is the position at which they occur in the list.

To be more specific, I am looking for something like this:

swapElementsAt :: Int -> Int -> [Int] -> [Int]

that would behave like that:

> swapElementsAt 1 3 [5,4,3,2,1] -- swap the first and third elements
[3,4,5,2,1]

I thought that a built-in function for this might exists in Haskell but I wasn't able to find it.

解决方案

Haskell doesn't have such a function, mainly because it is a little bit un-functional. What are you actually trying to achieve?

You can implement your own version of it (maybe there is a more idiomatic way to write this). Note that I assume that i < j, but it would be trivial to extend the function to correctly handle the other cases:

swapElementsAt :: Int -> Int -> [a] -> [a]
swapElementsAt i j xs = let elemI = xs !! i
                            elemJ = xs !! j
                            left = take i xs
                            middle = take (j - i - 1) (drop (i + 1) xs)
                            right = drop (j + 1) xs
                    in  left ++ [elemJ] ++ middle ++ [elemI] ++ right

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