Python拼接列表在向后工作时不包括第一个元素

Question

new_list = [i + 1 for i in range(16)]

sec_list = new_list[(len(new_list) - 2):0:-2]

print(new_list)

print(sec_list)

实际输出

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]

[15, 13, 11, 9, 7, 5, 3]

所需的输出：

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16]

[15, 13, 11, 9, 7, 5, 3, 1] <---- I want the 1 to be present

只是想弄清楚它是如何工作的。我以为[开始：结束：增加/减少]

Answer 1

这是您需要的

sec_list = new_list[(len(new_list) - 2)::-2]
print(sec_list) # [15, 13, 11, 9, 7, 5, 3, 1]