How to return the values of dictionary key that has gained new values in a different dictionary

Question

I have question a bit similar to:Replacing the value of a Python dictionary with the value of another dictionary that matches the other dictionaries key

However in my case I got two dictionaries

dict1 = {'foo' : ['val1' , 'val2' , 'val3'] , 'bar' : ['val4' , 'val5']}

dict2 = {'foo' : ['val2', 'val10', 'val11'] , 'bar' : ['val1' , 'val4']}

What I want to return is

dict3 = {'foo' : ['val10', 'val11'] , 'bar' : ['val1']}

and the opposite which is

dict4 = {'foo' : ['val1', 'val3'] , 'bar' : ['val5']}

where dict3 returns a dictionary of the values the keys 'foo' and 'bar' has gained in dict2 and the dict4 is a dictionary of the values the keys 'foo' and 'bar' has lost in dict2

One way I have tried solving this is to:

 iterate over both dictionaries then
    if key of dict1 == key of dict2
       return the values of the key in dict1 and compare with the values in dict2
       return the values that aren't in both 
       as a dictionary of the key and those values

This idea isn't working and obviously highly inefficient. I was hoping there is a more efficient working way to do this

Answer 1

Two dict comprehensions will do the trick:

dict3 = {k: [x for x in v if x not in dict1[k]] for k, v in dict2.items()}
print(dict3)
# {'foo': ['val10', 'val11'], 'bar': ['val1']}

dict4 = {k: [x for x in v if x not in dict2[k]] for k, v in dict1.items()}
print(dict4)
# {'foo': ['val1', 'val3'], 'bar': ['val5']}

The above two comprehensions basically filter out the values from key that don't exist in the other dictionary, which is done both ways.

You can also do this without dict comprehensions:

dict3 = {}
for k,v in dict2.items():
    dict3[k] = [x for x in v if x not in dict1[k]]

print(dict3)
# {'foo': ['val10', 'val11'], 'bar': ['val1']}

dict4 = {}
for k,v in dict1.items():
    dict4[k] = [x for x in v if x not in dict2[k]]

print(dict4)
# {'foo': ['val1', 'val3'], 'bar': ['val5']}