How to Get the Difference Between Two Dictionaries in Python?
I Have Two Dictionaries, and I Need to Find the Difference Between the Two, Which Should Give Me Both a Key and a Value. I Have Searched and Found Some...
I have two dictionaries, and I need to find the difference between the two, which should give me both a key and a value.
I have searched and found some addons/packages like datadiff and dictdiff-master, but when I try to import them in Python 2.7, it says that no such modules are defined.
I used a set here:
first_dict = {}
second_dict = {}
value = set(second_dict) - set(first_dict)
print value
My output is:
>>> set(['SCD-3547', 'SCD-3456'])
I am getting only keys, and I need to also get the values.
15 Answers
I think it's better to use the symmetric difference operation of sets to do that Here is the link to the doc.
>>> dict1 = {1:'donkey', 2:'chicken', 3:'dog'}
>>> dict2 = {1:'donkey', 2:'chimpansee', 4:'chicken'}
>>> set1 = set(dict1.items())
>>> set2 = set(dict2.items())
>>> set1 ^ set2
{(2, 'chimpansee'), (4, 'chicken'), (2, 'chicken'), (3, 'dog')}
It is symmetric because:
>>> set2 ^ set1
{(2, 'chimpansee'), (4, 'chicken'), (2, 'chicken'), (3, 'dog')}
This is not the case when using the difference operator.
>>> set1 - set2
{(2, 'chicken'), (3, 'dog')}
>>> set2 - set1
{(2, 'chimpansee'), (4, 'chicken')}
However it may not be a good idea to convert the resulting set to a dictionary because you may lose information:
>>> dict(set1 ^ set2)
{2: 'chicken', 3: 'dog', 4: 'chicken'}
Try the following snippet, using a dictionary comprehension:
value = { k : second_dict[k] for k in set(second_dict) - set(first_dict) }
In the above code we find the difference of the keys and then rebuild a dict taking the corresponding values.
Another solution would be dictdiffer ().
import dictdiffer
a_dict = {
'a': 'foo',
'b': 'bar',
'd': 'barfoo'
}
b_dict = {
'a': 'foo',
'b': 'BAR',
'c': 'foobar'
}
for diff in list(dictdiffer.diff(a_dict, b_dict)):
print diff
A diff is a tuple with the type of change, the changed value, and the path to the entry.
('change', 'b', ('bar', 'BAR'))
('add', '', [('c', 'foobar')])
('remove', '', [('d', 'barfoo')])
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'}
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()
}
Must Read
Example
print(diff_dicts({'a': 1, 'b': 1}, {'b': 2, 'c': 2}))
# {'c': (<class 'KeyError'>, 2), 'a': (1, <class 'KeyError'>), 'b': (1, 2)}