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]

另一种方法使用列表理解和abs

array = [0, 10, 20, 40]

reversed_array = [array[abs(indx)] for indx in range(abs(len(array)-1),1)]
            
reversed_array
[40, 20, 10, 0]

def reverse(text):
    output = []
    for i in range(len(text)-1, -1, -1):
        output.append(text[i])
    return output

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
>>>

如果你想在其他变量中存储反转列表的元素，那么你可以使用revArray = array[::-1]或revArray = list(reversed(array))。

但是第一个变体稍微快一点:

z = range(1000000)
startTimeTic = time.time()
y = z[::-1]
print("Time: %s s" % (time.time() - startTimeTic))

f = range(1000000)
startTimeTic = time.time()
g = list(reversed(f))
print("Time: %s s" % (time.time() - startTimeTic))

输出:

Time: 0.00489711761475 s
Time: 0.00609302520752 s

The cute Solution


class List_reverse():
    def reverse_list(self):
        list = [1, 2, 3, 4, 5, 6]
        list1 = []
        for i in range(1, len(list)+1):
            list1.append(list[-i])
        print(list1)


if __name__ == "__main__":
    obj = List_reverse()
    obj.reverse_list()
    enter code here