Python 如何对嵌套列表中的元素求和?
所以,我有一个类似这样的列表Python 如何对嵌套列表中的元素求和?,python,sum,nested-lists,Python,Sum,Nested Lists,所以,我有一个类似这样的列表 [[[1, 2], [3, 4], [5, 6]], [[7, 8], [9, 10], [11, 12]], [[13, 14], [15, 16], [17, 18]], [[19, 20], [21, 22], [23, 24]]] [[3, 7, 11], [15, 19, 23], [27, 31, 35], [39, 43, 27]] 我希望它看起来像这样 [[[1, 2], [3, 4], [5, 6]], [[7, 8], [9,
[[[1, 2], [3, 4], [5, 6]],
[[7, 8], [9, 10], [11, 12]],
[[13, 14], [15, 16], [17, 18]],
[[19, 20], [21, 22], [23, 24]]]
[[3, 7, 11],
[15, 19, 23],
[27, 31, 35],
[39, 43, 27]]
我希望它看起来像这样
[[[1, 2], [3, 4], [5, 6]],
[[7, 8], [9, 10], [11, 12]],
[[13, 14], [15, 16], [17, 18]],
[[19, 20], [21, 22], [23, 24]]]
[[3, 7, 11],
[15, 19, 23],
[27, 31, 35],
[39, 43, 27]]
也就是说,3=sum([1,2]),7=sum([3,4]),…
我尝试过为循环嵌套,但没有找到任何可以达到预期效果的东西,有人知道我如何做到这一点吗?这将非常好地完成这项工作,imo比列表理解更具可读性
lists = [[[1, 2], [3, 4], [5, 6]],
[[7, 8], [9, 10], [11, 12]],
[[13, 14], [15, 16], [17, 18]],
[[19, 20], [21, 22], [23, 24]]]
new_lists = []
for nested in lists:
new_ls = []
for ls in nested:
new_ls.append(sum(ls))
new_lists.append(new_ls)
>>> new_lists
[[3, 7, 11], [15, 19, 23], [27, 31, 35], [39, 43, 47]]
如果您愿意使用第三方库,可以使用NumPy和沿单个维度求和:
import numpy as np
A = np.array([[[1, 2], [3, 4], [5, 6]],
[[7, 8], [9, 10], [11, 12]],
[[13, 14], [15, 16], [17, 18]],
[[19, 20], [21, 22], [23, 24]]])
res = A.sum(2)
结果:
array([[ 3, 7, 11],
[15, 19, 23],
[27, 31, 35],
[39, 43, 47]])
另请参见:您也可以使用列表理解:
[[sum(x) for x in triple] for triple in lists]
在上面的列表理解中,triple将是您的三个Double列表,因此第一个for循环将涵盖这些。然后,x将是三元组中的每个双元组列表,因此我们对其求和,同时使用以下括号将其保留在原始三元组中:
[sum(x) for x in triple]
输出:
[[3, 7, 11], [15, 19, 23], [27, 31, 35], [39, 43, 47]]
请展示您对嵌套循环的尝试,以便我们解释为什么它没有给出预期的输出