反向返回列表

问题描述

我有这个问题:

编写一个函数reverse3(nums)，该函数接受一个长度为3的整数列表，称为nums，并返回一个具有相反顺序的元素的新列表，因此[1、2、3]变为[3， 2，1].

Write a function reverse3(nums) that takes a list of ints of length 3 called nums and returns a new list with the elements in reverse order, so [1, 2, 3] becomes [3, 2, 1].

我通过以下方式解决了该问题:

i solved it by:

def reverse3(nums):
    return [nums[2]] + [nums[1]] + [nums[0]]

但是，答案是直截了当的.我的主要问题是，当我不知道nums中有多少个整数时，如何反转nums?

however, the answer is straight foward. My main question, how do i get nums reversed, when i don't know how many ints there are in nums?.

我有这个:nums[::-1]确实返回nums反转.但是我正在寻找不同的方式.可能会循环吗?

I've got this:nums[::-1] which does return nums reversed.but i'm looking for a different way. probably looping?

我尝试过:

def reverse3(nums):
    return reversed(nums)

返回:<list_reverseiterator object at 0x10151ff90> #location?

which returns: <list_reverseiterator object at 0x10151ff90> #location?

推荐答案

返回一个迭代器.如果要获取列表对象，请使用 list :

reversed returns an iterator. If you want to get a list object, use list:

>>> def reverse3(nums):
...     return list(reversed(nums))
...
>>> reverse3([4,5,6])
[6, 5, 4]

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