intersection, union, difference, symmetric difference for dictionaries
Tim Chase
python.list at tim.thechases.com
Tue Feb 25 22:03:51 CET 2014
On 2014-02-25 14:40, Skip Montanaro wrote:
> What's the correct result of evaluating this expression?
>
> {'A': 1} | {'A': 2}
>
> I can see (at least) two possible "correct" answers.
I would propose at least four:
{'A': 1} # choose the LHS
{'A': 2} # choose the RHS
{'A': (1,2)} # a resulting pair of both
set(['A']) # you did set-ops, so you get a set
If dicts were to support set ops, the last one would be my preferred
result.
I just had to perform set operations on a pair of dicts earlier this
week, shrugged, and did things the manual/explicit way:
a_dict = dict(...)
b_dict = dict(...)
a_set = set(a_dict)
b_set = set(b_dict)
added_keys = b_set - a_set
removed_keys = a_set - b_set
same_keys = a_set & b_set
diff_keys = a_set ^ b_set
all_keys = a_set | b_set
It would save some space if I didn't have to duplicate all the keys
into sets (on the order of 10-100k small strings), instead being able
to directly perform the set-ops on the dicts. But otherwise, it was
pretty readable & straight-forward.
-tkc
More information about the Python-list
mailing list