How to get the difference between two dictionaries in Python?

Question

I have two dictionaries. I need to find the difference between the two which should give me both key and value.

I have searched and found some addons/packages like d

Answer 1

What about this? Not as pretty but explicit.

orig_dict = {'a' : 1, 'b' : 2}
new_dict = {'a' : 2, 'v' : 'hello', 'b' : 2}

updates = {}
for k2, v2 in new_dict.items():
    if k2 in orig_dict:    
        if v2 != orig_dict[k2]:
            updates.update({k2 : v2})
    else:
        updates.update({k2 : v2})

#test it
#value of 'a' was changed
#'v' is a completely new entry
assert all(k in updates for k in ['a', 'v'])

Answer 2

A function using the symmetric difference set operator, as mentioned in other answers, which preserves the origins of the values:

def diff_dicts(a, b, missing=KeyError):
    """
    Find keys and values which differ from `a` to `b` as a dict.

    If a value differs from `a` to `b` then the value in the returned dict will
    be: `(a_value, b_value)`. If either is missing then the token from 
    `missing` will be used instead.

    :param a: The from dict
    :param b: The to dict
    :param missing: A token used to indicate the dict did not include this key
    :return: A dict of keys to tuples with the matching value from a and b
    """
    return {
        key: (a.get(key, missing), b.get(key, missing))
        for key in dict(
            set(a.items()) ^ set(b.items())
        ).keys()
    }

Example

print(diff_dicts({'a': 1, 'b': 1}, {'b': 2, 'c': 2}))

# {'c': (<class 'KeyError'>, 2), 'a': (1, <class 'KeyError'>), 'b': (1, 2)}

How this works

We use the symmetric difference set operator on the tuples generated from taking items. This generates a set of distinct (key, value) tuples from the two dicts.

We then make a new dict from that to collapse the keys together and iterate over these. These are the only keys that have changed from one dict to the next.

We then compose a new dict using these keys with a tuple of the values from each dict substituting in our missing token when the key isn't present.

Answer 3

You were right to look at using a set, we just need to dig in a little deeper to get your method to work.

First, the example code:

test_1 = {"foo": "bar", "FOO": "BAR"}
test_2 = {"foo": "bar", "f00": "b@r"}

We can see right now that both dictionaries contain a similar key/value pair:

{"foo": "bar", ...}

Each dictionary also contains a completely different key value pair. But how do we detect the difference? Dictionaries don't support that. Instead, you'll want to use a set.

Here is how to turn each dictionary into a set we can use:

set_1 = set(test_1.items())
set_2 = set(test_2.items())

This returns a set containing a series of tuples. Each tuple represents one key/value pair from your dictionary.

Now, to find the difference between set_1 and set_2:

print set_1 - set_2
>>> {('FOO', 'BAR')}

Want a dictionary back? Easy, just:

dict(set_1 - set_2)
>>> {'FOO': 'BAR'}