如何比较具有不同键但相似值的两个字典并删除重复项

Question

I am quite new to python and I am keen on learning. I have two dictionaries that have different keys but similar values in it, as follows:

dict_a = {'r1': ['c5', 'c6', 'c7', 'c8'], 'r2': ['c9', 'c10', 'c11'], 'r3': ['c12', 'c13', 'c14', 'c15']}

dict_b = {'f1': ['c1', 'c2', 'c3', 'c4', 'c5', 'c6', 'c7', 'c8', 'c9', 'c10', 'c11', 'c12', 'c13', 'c14', 'c15']}

Is it possible to compare both dictionaries and delete the duplicate values in it? I would like to obtain the following dictionaries at the end:

dict_a_new = {'r1': ['c5', 'c6', 'c7', 'c8'], 'r2': ['c9', 'c10', 'c11'], 'r3': ['c12', 'c13', 'c14', 'c15']}

dict_b_new = {'f1': ['c1', 'c2', 'c3', 'c4'}

I have tried the following syntax, but it doesn't work for me.

dict_b_new = {k: dict_b[k] for k in set(dict_b) - set(dict_a)}

Any suggestions will be highly appreciated.

Answer 1

You can create a set with all the items in dict_a , and then create dict_b_new with all the items in dict_b that are not in the set:

set_a = {item for lst in dict_a.values() for item in lst}
dict_b_new = {k: [item for item in lst if item not in set_a] for k, lst in dict_b.items()}

Demo: https://replit.com/@blhsing/PuzzledAgreeableProspect

Answer 2

I suppose dict_a has multiple keys, not only r1 , r2 , r3 , and dict_b has a single key.

from itertools import chain

difference = set(*dict_b.values()) - set(chain(*dict_a.values()))
# {'c1', 'c2', 'c3', 'c4'}

dict_b_new = {k: difference for k in dict_b.keys()}

itertools.chain concatenate iterables.

Ref: https://docs.python.org/3/library/itertools.html#itertools.chain