Intersecting two dictionaries

A little known fact is that you don't need to construct sets to do this:

In Python 2:

In [78]: d1 = {'a': 1, 'b': 2}In [79]: d2 = {'b': 2, 'c': 3}In [80]: d1.viewkeys() & d2.viewkeys()Out[80]: {'b'}

In Python 3 replace viewkeys with keys; the same applies to viewvalues and viewitems.

From the documentation of viewitems:

In [113]: d1.viewitems??Type:       builtin_function_or_methodString Form:<built-in method viewitems of dict object at 0x64a61b0>Docstring:  D.viewitems() -> a set-like object providing a view on D's items

For larger dicts this also slightly faster than constructing sets and then intersecting them:

In [122]: d1 = {i: rand() for i in range(10000)}In [123]: d2 = {i: rand() for i in range(10000)}In [124]: timeit d1.viewkeys() & d2.viewkeys()1000 loops, best of 3: 714 µs per loopIn [125]: %%timeits1 = set(d1)s2 = set(d2)res = s1 & s21000 loops, best of 3: 805 µs per loopFor smaller `dict`s `set` construction is faster:In [126]: d1 = {'a': 1, 'b': 2}In [127]: d2 = {'b': 2, 'c': 3}In [128]: timeit d1.viewkeys() & d2.viewkeys()1000000 loops, best of 3: 591 ns per loopIn [129]: %%timeits1 = set(d1)s2 = set(d2)res = s1 & s21000000 loops, best of 3: 477 ns per loop

We're comparing nanoseconds here, which may or may not matter to you. In any case, you get back a set, so using viewkeys/keys eliminates a bit of clutter.

In general, to construct the intersection of dictionaries in Python, you can first use the & operator to calculate the intersection of sets of the dictionary keys (dictionary keys are set-like objects in Python 3):

dict_a = {"a": 1, "b": 2}dict_b = {"a": 2, "c": 3} intersection = dict_a.keys() & dict_b.keys()  # {'a'}

On Python 2 you have to convert the dictionary keys to sets yourself:

keys_a = set(dict_a.keys())keys_b = set(dict_b.keys())intersection = keys_a & keys_b

Then given the intersection of the keys, you can then build the intersection of your values however is desired. You have to make a choice here, since the concept of set intersection doesn't tell you what to do if the associated values differ. (This is presumably why the & intersection operator is not defined directly for dictionaries in Python).

In this case it sounds like your values for the same key would be equal, so you can just choose the value from one of the dictionaries:

dict_of_dicts_a = {"a": {"x":1}, "b": {"y":3}}dict_of_dicts_b = {"a": {"x":1}, "c": {"z":4}} shared_keys = dict_of_dicts_a.keys() & dict_of_dicts_b.keys()# values equal so choose values from a:dict_intersection = {k: dict_of_dicts_a[k] for k in shared_keys }  # {"a":{"x":1}}

Other reasonable methods of combining values would depend on the types of the values in your dictionaries, and what they represent. For example you might also want the union of values for shared keys of dictionaries of dictionaries. Since the union of dictionaries doesn't depend on the values, it is well defined, and in python you can get it using the | operator:

# union of values for each key in the intersection:dict_intersection_2 = { k: dict_of_dicts_a[k] | dict_of_dicts_b[k] for k in shared_keys }

Which in this case, with identical dictionary values for key "a" in both, would be the same result.

In [1]: d1 = {'a':1, 'b':4, 'f':3}In [2]: d2 = {'a':1, 'b':4, 'd':2}In [3]: d = {x:d1[x] for x in d1 if x in d2}In [4]: dOut[4]: {'a': 1, 'b': 4}