(PYTHON) 如何在列表中添加每第 N 项元素以生成新列表？

Question

假设我们有以下列表

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]

现在我想将每 3 个数字相加以提供 6 个列表的长度，

[6, 15, 24, 33, 42, 51]

我想在 python 中执行此操作...请帮忙！（我的问题措辞很奇怪吗？）

直到现在我尝试过

z = np.zeros(6)
p = 0
cc = 0
for i in range(len(that_list)):
    p += that_list[i]
    cc += 1
    if cc == 3:
       t = int((i+1)/3)
       z[t] = p
       cc = 0
       p = 0

但它没有用......

Answer 1

考虑使用列表理解：

>>> nums = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]
>>> [sum(nums[i:i+3]) for i in range(0, len(nums), 3)] 
[6, 15, 24, 33, 42, 51]

或麻木：

>>> import numpy as np
>>> nums = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]
>>> np.add.reduceat(nums, np.arange(0, len(nums), 3))
>>> array([ 6, 15, 24, 33, 42, 51])

如果由于某种原因需要使用手动循环：

nums = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18]
result = []
group_size, group_sum, group_length = 3, 0, 0
for num in nums:
    group_sum += num
    group_length += 1
    if group_length == group_size:
        result.append(group_sum)
        group_sum, group_length = 0, 0
print(result)  # [6, 15, 24, 33, 42, 51]