从2个列表中收集唯一值

Question

I currently have 2 "set" lists, both contain mostly the same values ie:

a = [1,2,3,4,5,6]

b = [1,2,3,4,5,6,7,8]

What I'm trying to do is compare the two and only return 7 & 8 into another list however I'm having no luck, I've tried a few methods I've found on here such as:

c = [item for item in a if item not in b]

However I've had no luck does anyone know a quick and fairly easy method of doing this? Both lists have been "set" previously to remove any duplicates within their own lists also (just felt I should add that if that makes a difference)

Just to be clear, as I think I wasn't sorry, the example values are already in set format, in my head for some reason when you used set on a list it acted like the array_unique PHP function.

Thanks.

Answer 1

You can use sets:

set(b) - set(a)
Out[65]: {7, 8}

Or more explicitly:

set(b).difference(a)
Out[67]: {7, 8}

For symmetric difference, you can use ^ . It will return elements that are in a or b, but not in both. Assume a has an additional element, 9:

a = [1, 2, 3, 4, 5, 6, 9]
b = [1, 2, 3, 4, 5, 6, 7, 8]
set(b) ^ set(a)
Out[70]: {7, 8, 9}

Or,

set(b).symmetric_difference(a)
Out[71]: {7, 8, 9}

If they are already sets, you can just do b - a or b ^ a .

Your attempt does not work because there is no item in a which is not in b . For that difference ( ba ), you need: [item for item in b if item not in a] .

Answer 2

Another approach is using a defaultdict to count the number of occurrences of each element from both lists:

from collections import defaultdict
from itertools import chain

a = [1,2,3,4,5,6,9]

b = [1,2,3,4,5,6,7,8,9]

d = defaultdict(int)

for num in chain(a, b):
    d[num] += 1


print([key for key, value in d.items() if value == 1])

Output:

[7, 8]

Answer 3

list(set(a)-set(b))+list(set(b)-set(a))