通过切换相反索引的引用来就地反转:

>>> l = [1,2,3,4,5,6,7]    
>>> for i in range(len(l)//2):
...     l[i], l[-1-i] = l[-1-i], l[i]
...
>>> l
[7, 6, 5, 4, 3, 2, 1]

内置功能最少，假设是面试设置

array = [1, 2, 3, 4, 5, 6,7, 8]
inverse = [] #create container for inverse array
length = len(array)  #to iterate later, returns 8 
counter = length - 1  #because the 8th element is on position 7 (as python starts from 0)

for i in range(length): 
   inverse.append(array[counter])
   counter -= 1
print(inverse)

>>> L = [1, 2, 3, 4]
>>> L = [L[-i] for i in range(1, len(L) + 1)]
>>> L
[4, 3, 2, 1]

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

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

这里将解释扩展片语法。请参见文档。

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