如何删除Haskell列表中的第n个元素 [英] how do you delete the nth element in a list in Haskell

问题描述

如何编写函数的实现，该函数采用数字n和列表，并从列表中删除位置n的元素.例如:删除0 [1、2、3、4] = [2、3、4]

How to write the implementation for the function which takes a number n and a list and removes the element at position n from the list. For example: remove 0 [1, 2, 3, 4] = [2, 3, 4]

这是我的尝试:

import Data.List.Split
remove :: Int -> [a] -> [a]
remove xs = let remove n xs = let (as, bs) = splitAt n xs in as ++ tail bs

但是它不起作用，并向我显示变量不在范围内" .有谁知道如何修理它?谢谢！

But it doesn't work and shows me "Variable not in scope". Does anyone know how to fix it? Appreciate!!

推荐答案

我写了一种非常直观的递归方法:

I wrote a pretty intuitive recursive approach:

remove :: Int -> [a] -> [a]
remove _ [] = []
remove 0 (x:xs) = xs
remove n (x:xs) = x : remove (n-1) (xs)

考虑到我正在考虑第一个元素为索引0.如果您有任何疑问，请告诉我.

Take into account that I am considering the first element to be index 0. Let me know if you have any doubts.

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