find the non intersection between 2 list, including duplicate

Question

I tried to look up this question before asking but could not find any satisfactory. so I have 2 list like this

a=[1,2,3,4,4]
b=[1,1,2,3,4]

I tried this:

set(a) - set(b)

but got this

set()

what I want is this

[1,4]

Since set a has 2 4s and set b has 2 1s. What can I do? Thank you!

*"set a has 2 4s"* - No, it doesn't, sets don't have duplicates, as soon as you use `set`, set a has one 4, likewise for set b with 1s — Nick is tired, Oct 03 '19 at 12:21
But even if they were multisets (math term), then `set(a) - set(b)` would only have one element, 4. Please explain what you want because your output doesn't match your attempts. — h4z3, Oct 03 '19 at 12:24

score 2 · Answer 1 · answered Oct 03 '19 at 12:25

With collections.Counter object to compare occurrences of respective values:

from collections import Counter

a = [1, 2, 3, 4, 4]
b = [1, 1, 2, 3, 4]
a_counts, b_counts = Counter(a), Counter(b)
res = [a_key for a_key, b_key in zip(a_counts, b_counts)
       if a_counts[a_key] != b_counts[b_key]]

print(res)   # [1, 4]

Dani Mesejo · Accepted Answer · 2019-10-03T12:35:55.680

Use collections.Counter, is the Python implementation of a multiset:

from collections import Counter

a = [1, 2, 3, 4, 4]
b = [1, 1, 2, 3, 4]
c = [1, 1, 1, 2, 3, 4]

counts_a = Counter(a)
counts_b = Counter(b)
counts_c = Counter(c)

result_b = (counts_a | counts_b) - (counts_a & counts_b)
result_c = (counts_a | counts_c) - (counts_a & counts_c)

print(list(result_b.elements()))
print(list(result_c.elements()))

Output

[1, 4]
[1, 1, 4]

Note that (counts_a | counts_b) - (counts_a & counts_b) is the Python equivalent of the mathematical formula.

find the non intersection between 2 list, including duplicate

2 Answers2