Compare two large dictionaries and create lists of values for keys they have in common

Question

I have a two dictionaries like:

dict1 = { (1,2) : 2, (2,3): 3, (1,3): 3}
dict2 = { (1,2) : 1, (1,3): 2}

What I want as output is two list of values for the items which exist in both dictionaries:

[2,3]
[1,2]

What I am doing right now is something like this:

list1 = []
list2 = []

for key in dict1.keys():
    if key in dict2.keys():
        list1.append(dict1.get(key))
        list2.append(dict2.get(key))

This code is taking too long running which is not something that I am looking forward to. I was wondering if there might be a more efficient way of doing it?

Answer 1

commons = set(dict1).intersection(set(dict2))
list1 = [dict1[k] for k in commons]
list2 = [dict2[k] for k in commons]

Answer 2

Don't use dict.keys. On python2.x, it creates a new list every time it is called (which is an O(N) operation -- And list.__contains__ is another O(N) operation on average). Just rely on the fact that dictionaries are iterable containers directly (with O(1) lookup):

list1 = []
list2 = []

for key in dict1:
    if key in dict2:
        list1.append(dict1.get(key))
        list2.append(dict2.get(key))

---

Note that on python2.7, you can use viewkeys to get the intersection directly:

>>> a = {'foo': 'bar', 'baz': 'qux'}
>>> b = {'foo': 'bar'}
>>> a.viewkeys() & b
set(['foo'])

(on python3.x, you can use keys here instead of viewkeys)

for key in dict1.viewkeys() & dict2:
    list1.append(dict1[key]))
    list2.append(dict2[key]))