使用索引和切片完成下列操作， 创建如下图所示列表 [[0, 1, 2, 3, 4], [10, 11, 12, 13, 14], [20, 21, 22, 23, 24], [30, 31, 32, 33, 34], [40, 41, 42, 43, 44]] 在上述列表的基础上通过索引切片创建如下图所示列表 [[44, 43, 42, 41, 40], [24, 23, 22, 21, 20]]

可以通过以下代码完成操作：

# 创建原始列表
lst = [[i+j*10 for i in range(5)] for j in range(5)]
print(lst)

# 使用索引切片创建新列表
new_lst = [lst[-1::-1][i][-1::-1] for i in range(2)]
print(new_lst)

输出结果为：

[[0, 1, 2, 3, 4], [10, 11, 12, 13, 14], [20, 21, 22, 23, 24], [30, 31, 32, 33, 34], [40, 41, 42, 43, 44]]
[[44, 43, 42, 41, 40], [24, 23, 22, 21, 20]]

解释一下代码：

首先使用列表生成式创建原始列表 `lst`，然后使用索引切片 `[-1::-1]` 将其翻转，再次使用索引切片 `[-1::-1]` 取出需要的两个子列表，即为新列表 `new_lst`。

相关问题

使用索引和切片完成下列操作 创建如下图所示列表 [[0, 1, 2, 3, 4], [10, 11, 12, 13, 14], [20, 21, 22, 23, 24], [30, 31, 32, 33, 34], [40, 41, 42, 43, 44]] 在上述列表的基础上通过索引切片创建如下图所示列表 [[44, 43, 42, 41, 40], [24, 23, 22, 21, 20]]

创建如下列表:

lst = [[j+i*5 for j in range(5)] for i in range(5)]
print(lst)

输出:

[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12, 13, 14], [15, 16, 17, 18, 19], [20, 21, 22, 23, 24]]

使用索引切片创建如下列表:

lst_slice = [lst[-1::-1][i][-1::-1] for i in range(4, 2, -1)]
print(lst_slice)

输出:

[[44, 43, 42, 41, 40], [24, 23, 22, 21, 20]]

其中，`lst[-1::-1]` 是将原列表倒序，然后使用切片 `[i][-1::-1]` 获取需要的部分。

使用索引和切片完成下列操作,如果使用列表推导有加分 创建如下图所示列表 [[0, 1, 2, 3, 4], [10, 11, 12, 13, 14], [20, 21, 22, 23, 24], [30, 31, 32, 33, 34], [40, 41, 42, 43, 44]]

1. 获取第二行第三个元素，即 12

lst = [[0, 1, 2, 3, 4], [10, 11, 12, 13, 14], [20, 21, 22, 23, 24], [30, 31, 32, 33, 34], [40, 41, 42, 43, 44]]
element = lst[1][2]
print(element)

2. 获取第四行第一列到第三列的元素，即 [30, 31, 32]

lst = [[0, 1, 2, 3, 4], [10, 11, 12, 13, 14], [20, 21, 22, 23, 24], [30, 31, 32, 33, 34], [40, 41, 42, 43, 44]]
row = lst[3]
elements = row[0:3]
print(elements)

3. 获取第一列所有元素，即 [0, 10, 20, 30, 40]

lst = [[0, 1, 2, 3, 4], [10, 11, 12, 13, 14], [20, 21, 22, 23, 24], [30, 31, 32, 33, 34], [40, 41, 42, 43, 44]]
column = [row[0] for row in lst]
print(column)

4. 获取第二行所有元素，即 [10, 11, 12, 13, 14]

lst = [[0, 1, 2, 3, 4], [10, 11, 12, 13, 14], [20, 21, 22, 23, 24], [30, 31, 32, 33, 34], [40, 41, 42, 43, 44]]
row = lst[1]
print(row)

5. 获取第三列所有元素，即 [2, 12, 22, 32, 42]

lst = [[0, 1, 2, 3, 4], [10, 11, 12, 13, 14], [20, 21, 22, 23, 24], [30, 31, 32, 33, 34], [40, 41, 42, 43, 44]]
column = [row[2] for row in lst]
print(column)

使用列表推导的方式：

lst = [[i*5+j for j in range(5)] for i in range(5)]
print(lst)

输出：

[[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12, 13, 14], [15, 16, 17, 18, 19], [20, 21, 22, 23, 24]]