如何对两个迭代器“列表”中的值求和?

您可以使用此函数对每个值求和,而不必将所有内容加载到内存中:

def sumIter(iter):
    result = [0, 0, 0, 0, 0, 0] #Assuming there are always 6 items in each sub-list
    for list in iter:
       result = [(result[i] + list[i]) for i in range(6)]

然后:

sum1 = sumIter(iter1)
sum2 = sumIter(iter2)
result = [(sum1[i] + sum2[i]) for i in range(6)]

使用 zip

l1 = [
    [1, 2, 3, 4, 5, 6],
    [2, 4, 6, 8, 10, 12],
    [3, 5, 8, 6, 1, 19],
    [5, 9, 1, 9, 4, 6]
]
l2 = [
    [5, 4, 3, 2, 1, 9],
    [6, 3, 8, 1, 1, 6],
    [1, 3, 2, 8, 2, 3],
    [5, 2, 9, 4, 2, 5]
]


l1 = (sum(i) for i in zip(*l1))
l2 = (sum(i) for i in zip(*l2))

print( [sum(i) for i in zip(l1, l2)] )

输出:

[28, 32, 40, 42, 26, 66]

reduce 因为行可以添加到numpy数组中。

减少 是python2中的内置函数

import numpy as np
from functools import reduce # only in python3

def sumup(one_row, another_row):
   return one_row + another_row

test_list = np.array([[1, 2, 3, 4, 5, 6],
                   [2, 4, 6, 8, 10, 12],
                   [3, 5, 8, 6, 1, 19], 
                   [5, 9, 1, 9, 4, 6]])

reduce(sumup, test_list)

输出

array([11, 20, 18, 27, 20, 43])

使用 numpy.sum

import numpy as np

l1 = np.sum([[1, 2, 3, 4, 5, 6], [2, 4, 6, 8, 10, 12], [3, 5, 8, 6, 1, 19], [5, 9, 1, 9, 4, 6]], axis=0)
l2 = np.sum([[5, 4, 3, 2, 1, 9],[6, 3, 8, 1, 1, 6], [1, 3, 2, 8, 2, 3],[5, 2, 9, 4, 2, 5]], axis=0)

print(l1 + l2)

[28 32 40 42 26 66]