在python中将列表分组在一起

Question

我正在尝试将self.L列表分为五个组，然后反转两个单独的组，我希望我的输出看起来如下

    step1.(group the  list into two section of five)

    [[1,3,5,67,8],[90,100,45,67,865]]

    step2.(reverse the numbers within the two groups)

    [[8,67,5,3,1],[865,67,45,100,90]]

我有的代码

        class Flip():
    def __init__(self,answer):
        self.Answer = answer
        self.matrix = None
        self.L = [1,3,5,67,8,90,100,45,67,865,]


    def flip(self):
        if self.Answer == 'h':
            def grouping(self):
                for i in range(0, len(self.L), 5):
                   self.matrix.append(self.L[i:i+5])
            print(self.matrix)

if __name__ =='__main__':
    ans = input("Type h for horizontal flip")
    work = Flip(ans)
    work.flip()

当我运行代码时，我的输出是

    None

Answer 1

要创建每个包含5个元素的嵌套列表，可以使用列表理解 ：

lst = [1,3,5,67,8,90,100,45,67,865]
new_list = [lst[i:i+5] for i in range(0, len(lst), 5)]

然后，要反转嵌套列表中值的顺序，可以使用.reverse()

for elem in new_list:
    elem.reverse()
print (new_list)

输出：

[[8, 67, 5, 3, 1], [865, 67, 45, 100, 90]]

Answer 2

假设我们在变量L收到初始列表：

L = [1,3,5,67,8,90,100,45,67,865]

L1 = L[0:5]
L2 = L[5:10]
L=[L1,L2]
print (L)
L1.reverse()
L2.reverse()
L=[L1,L2]
print (L)