可以使用__reverse__来完成，它会返回一个生成器。

>>> l = [1,2,3,4,5]
>>> for i in l.__reversed__():
...   print i
... 
5
4
3
2
1
>>>

使用列表。Reverse将列表就地反转:

>>> xs = [0, 10, 20, 40]
>>> xs.reverse()
>>> xs
[40, 20, 10, 0]

使用切片以相反的顺序创建一个新列表:

>>> xs[::-1]
[40, 20, 10, 0]

要获得一个新的反向列表，应用reversed函数并将项目收集到一个列表中:

>>> xs = [0, 10, 20, 40]
>>> list(reversed(xs))
[40, 20, 10, 0]

在列表中向后迭代:

>>> xs = [0, 10, 20, 40]
>>> for x in reversed(xs):
...     print(x)
40
20
10
0

可以使用__reverse__来完成，它会返回一个生成器。

>>> l = [1,2,3,4,5]
>>> for i in l.__reversed__():
...   print i
... 
5
4
3
2
1
>>>

这个类使用Python魔法方法和迭代器进行反转，并反转列表:

class Reverse(object):
    """ Builds a reverse method using magic methods """

    def __init__(self, data):
        self.data = data
        self.index = len(data)


    def __iter__(self):
        return self

    def __next__(self):
        if self.index == 0:
            raise StopIteration

        self.index = self.index - 1
        return self.data[self.index]


REV_INSTANCE = Reverse([0, 10, 20, 40])

iter(REV_INSTANCE)

rev_list = []
for i in REV_INSTANCE:
    rev_list.append(i)

print(rev_list)  

输出

[40, 20, 10, 0]

我在一次面试的python代码测试中遇到了这个问题。 以下是我的答案。 注意它适用于任何值任何长度

def get_reverse(list_check, count_num):
    final_list =[]
    for index in range(list_length):
        value = list_check[count_num]
        final_list.append(value)
        count_num = count_num -1

    return final_list

new_list = ['A', 'GOAT', 'C', 'D', 'Mac']

list_length = len(new_list)
x = list_length -1

print(get_reverse(new_list, x))