如何从存储在元组列表中的两个元素元组中生成两个列表

Question

我有一个包含许多列表的列表，其中有4个元组

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

我希望将它们分为两个单独的列表，如：

my_list2 = [12,10,4,2,110,34,12,55]
my_list3 = [1,3,0,0,1,2,1,3]

我的尝试是使用map函数进行此操作

my_list2 , my_list3 = map(list, zip(*my_list))

但这给了我一个错误：

ValueError: too many values to unpack (expected 2)

Answer 1

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]

Answer 2

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]

Answer 3

您的方法非常接近，但您需要首先平坦化：

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]