Haskell替换列表中的元素 [英] Haskell replace element in list

问题描述

示例：

是否有内置函数来替换haskell中给定索引处的元素？

replaceAtIndex（2，foo，[bar，bar，bar]）

应该给：

  [bar，bar，foo] 
  

我知道我可以创建自己的函数，但它似乎应该是内置的。

解决方案

如果您需要更新特定索引处的元素，则列表不是最好的数据结构。您可能要考虑使用 Seq 。 htmlrel =noreferrer> Data.Sequence ，在这种情况下，您要查找的函数是 update :: Int - > a - > Seq a - > Seq a 。

 > import Data.Sequence 
>更新2foo$ fromList [bar，bar，bar] 
 fromList [bar，bar，foo] 
  

Is there any built-in function to replace an element at a given index in haskell?

Example:

replaceAtIndex(2,"foo",["bar","bar","bar"])

Should give:

["bar", "bar", "foo"]

I know i could make my own function, but it just seems it should be built-in.

解决方案

If you need to update elements at a specific index, lists aren't the best data structure for that. You might want to consider using Seq from Data.Sequence instead, in which case the function you're looking for is update :: Int -> a -> Seq a -> Seq a.

> import Data.Sequence
> update 2 "foo" $ fromList ["bar", "bar", "bar"]
fromList ["bar","bar","foo"]

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