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)

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))

下面是一种使用生成器对反向进行惰性计算的方法:

def reverse(seq):
    for x in range(len(seq), -1, -1): #Iterate through a sequence starting from -1 and increasing by -1.
        yield seq[x] #Yield a value to the generator

现在像这样迭代:

for x in reverse([1, 2, 3]):
    print(x)

如果你需要一个列表:

l = list(reverse([1, 2, 3]))

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

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

要反转相同的列表，请使用:

array.reverse()

将反向列表赋值到其他列表使用:

newArray = array[::-1]