如何确定列表之间的差异？

Question

I have two lists:

first_list = [('A', 'Name A'), ('B', 'Name B'), ('C', 'Other Name')]
second_list = [('A', 'Name A'), ('B', 'Name B'), ('C', 'Name C'), ('D', 'Name D')]

I want such list: third_list = [('D', 'Name D')]

I used: third_list = list(set(second_list) ^ set(first_list)) but it return me: third_list = [('C', 'Name C'), ('D', 'Name D')] .

So as you can see I want list where first items of tuples are different. ('C', 'Other Name') and ('C', 'Name C') must be the same cause there first items in tuples are the same.

Answer 1

Check online demo

first_list = [('A', 'Name A'), ('B', 'Name B'), ('C', 'Other Name')]
second_list = [('A', 'Name A'), ('B', 'Name B'), ('C', 'Name C'), ('D', 'Name D')]

first_dict = dict(first_list)
second_dict = dict(second_list)
value = { k : second_dict[k] for k in set(second_dict) - set(first_dict) }
print(value)

Answer 2

This will do the job:

first_second = [('A', 'Name A'), ('B', 'Name B'), ('C', 'Other Name')]
second_second = [('A', 'Name A'), ('B', 'Name B'), ('C', 'Name C'), ('D', 'Name D')]
d_one=dict(first_second)
d_two=dict(second_second)
res=[(i,j) for i,j in d_two.items() if i in set(d_two).symmetric_difference(set(d_one)) ]

Then res is

[('D', 'Name D')]

Answer 3

[x for x in second_list if x[0] not in dict(first_list)]

[('D', 'Name D')]