比较元组和列表的字典

Question

I have found multiple posts (this one too) that was about this topic but none of those options would work for my dictionaries. I wanna compare my dictionnaries and know the number of values identical and also the pair of key-values unique to each dictionaries.

I am working with two dictionaries with Tuples as key and list as value (where the second value is another list) as follow:

Dict1:{(10, 11): ['C', ['T']],
       (20, 21): ['C', ['T']],
       (34, 35): ['G', ['A']],
       (68, 69): ['A', ['T','G']]}


Dict2:{(10, 11): ['C', ['T']],
       (20, 21): ['C', ['A']],
       (40, 41): ['T', ['G']],
       (68, 69): ['A', ['T','G']]}

and I would like to compare those dictionnary and have different output. Using my example that's the variable I would like to have:

• 2 values are identical and present in both dict
• 2 values are only in dict1
• 2 values are only in dict2

I was about to loop over dict1 and compare each key to all dict2 each time (and having variables that i'll updates each time a condition is met) but I am aware that it is probably not the most efficient way of doing it.

Does anyone have a quicker idea?

Thanks

Answer 1

You can first filter out the keys using set-like methods of dict.keys() objects , and then proceed to get the pairs:

>>> same_keys = Dict1.keys() & Dict2.keys()
>>> dict1_unq_keys = Dict1.keys() - Dict2.keys()
>>> dict2_unq_keys = Dict2.keys() - Dict1.keys()
>>> same_pairs = [(key, Dict1[key]) for key in same_keys if Dict1[key] == Dict2[key]]
>>> Dict1_unq_pair = [(key, Dict1[key]) for key in dict1_unq_keys]
>>> Dict2_unq_pair = [(key, Dict2[key]) for key in dict2_unq_keys]

>>> same_pairs
[((68, 69), ['A', ['T', 'G']]), ((10, 11), ['C', ['T']])]

>>> Dict1_unq_pair
[((34, 35), ['G', ['A']])]

>>> Dict2_unq_pair
[((40, 41), ['T', ['G']])]

Note:

If it is possible for you to use tuples as dict values instead of lists , this could be done more easily, by directly using dict.items() set operations to get the same pairs.

For example, if Dict1 and Dict2 were of the following form:

>>> Dict1
{(10, 11): ('C', ('T',)),
 (20, 21): ('C', ('T',)),
 (34, 35): ('G', ('A',)),
 (68, 69): ('A', ('T', 'G'))}

>>> Dict2
{(10, 11): ('C', ('T',)),
 (20, 21): ('C', ('A',)),
 (40, 41): ('T', ('G',)),
 (68, 69): ('A', ('T', 'G'))}

# Then you could simple do:
>>> same_pairs = list(Dict1.items() & Dict2.items())

>>> same_pairs
[((68, 69), ('A', ('T', 'G'))), ((10, 11), ('C', ('T',)))]