你的方法是相当接近，但你需要先变平：

from itertools import chain

my_list = [[(12, 1), (10, 3), (4, 0), (2, 0)], [(110, 1), (34, 2), (12, 1), (55, 3)]]

my_list2 , my_list3 = map(list,zip(*chain.from_iterable(my_list)))

my_list2
# [12, 10, 4, 2, 110, 34, 12, 55]

my_list3
# [1, 3, 0, 0, 1, 2, 1, 3]

另一种简单的方法：

my_list = [[(12, 1), (10, 3), (4, 0), (2, 0)], [(110, 1), (34, 2), (12, 1), (55, 3)]]

first = []
second = []

for inner in my_list:
    for each in inner:
        first.append(each[0])
        second.append(each[1])

print(first)  # [12, 10, 4, 2, 110, 34, 12, 55]
print(second)  # [1, 3, 0, 0, 1, 2, 1, 3]

试试看：

my_list = [[(12, 1), (10, 3), (4, 0), (2, 0)], [(110, 1), (34, 2), (12, 1), (55, 3)]]

my_list2, my_list3 = map(list, zip(*[j for i in my_list for j in i]))
print(my_list2)
# [12, 10, 4, 2, 110, 34, 12, 55]
print(my_list3)
# [1, 3, 0, 0, 1, 2, 1, 3]

你可以使用列表解析（5.1.3）。
元组的第一个数量：

my_list2 = [tuple[0] for inner in my_list for tuple in inner]

元组的第二个数量：

my_list3 = [tuple[1] for inner in my_list for tuple in inner]

这个怎么样？

my_list = [[(12, 1), (10, 3), (4, 0), (2, 0)], [(110, 1), (34, 2), (12, 1), (55, 3)]]

flatten = lambda l: [item for my_list in l for item in my_list]

list1, list2 = zip(*flatten(my_list))

这里有一种方法：

my_list = [[(12, 1), (10, 3), (4, 0), (2, 0)], [(110, 1), (34, 2), (12, 1), (55, 3)]]

my_list2 = [a for b in [[t[0] for t in my_list[i]] for i,n in enumerate(my_list)] for a in b]
my_list3 = [a for b in [[t[1] for t in my_list[i]] for i,n in enumerate(my_list)] for a in b]

print(my_list2)
print(my_list3)

输出：

[12, 10, 4, 2, 110, 34, 12, 55]
[1, 3, 0, 0, 1, 2, 1, 3]

最低限度的解决方案如何：

my_list = [[(12, 1), (10, 3), (4, 0), (2, 0)], [(110, 1), (34, 2), (12, 1), (55, 3)]]

my_list2, my_list3 = zip(*sum(my_list, []))

print(my_list2)
print(my_list3)

输出

> python3 test.py
(12, 10, 4, 2, 110, 34, 12, 55)
(1, 3, 0, 0, 1, 2, 1, 3)
>

将元组列表转换为2d数组，然后对数组进行切片

my_list = [[(12, 1), (10, 3), (4, 0), (2, 0)],
       [(110, 1), (34, 2), (12, 1), (55, 3)]]

a = np.array(my_list)
my_list2= np.array([])
my_list3= np.array([])
my_list2=np.append(my_list2,[item[:,0] for item in a])
my_list3=np.append(my_list3,[item[:,1] for item in a])
print(my_list2)
print(my_list3)

输出：

[ 12.  10.   4.   2. 110.  34.  12.  55.]
[1. 3. 0. 0. 1. 2. 1. 3.]