How to compare two lists of dicts for multiple key, value pairs?

Question

I have two list s of dict s, one is a modified subset of the other. I would like to get the elements of list_one that don't appear in list_two , based on two keys. Example:

list_one = [{'name': 'alf', 'age': 25},
            {'name': 'alf', 'age': 50},
            {'name': 'cid', 'age': 30}]
list_two = [{'name': 'alf', 'age': 25, 'hair_color': 'brown'},
            {'name': 'cid', 'age': 30, 'hair_color': 'black'}]
desired_list = [{'name': 'alf', 'age': 50}]

How can I accomplish this? I have a feeling it is with some sort of list comprehension , as such:

desired_list = [x for x in list_one if x['name'] != x2['name'] and x['age'] != x2['age'] 
                for all x2 in list_two]

Answer 1

I think this is easily done with two comprehensions as:

Code:

have_2 = {(d['name'], d['age']) for d in list_two}
extra = [d for d in list_one if (d['name'], d['age']) not in have_2]

This first creates a set of tuple s which we already have, then checks which dict s do not match any of these existing keys.

Test Code:

list_one = [{'name': 'alf', 'age': 25},
            {'name': 'alf', 'age': 50},
            {'name': 'cid', 'age': 30}]
list_two = [{'name': 'alf', 'age': 25, 'hair_color': 'brown'},
            {'name': 'cid', 'age': 30, 'hair_color': 'black'}]

have_2 = {(d['name'], d['age']) for d in list_two}
extra = [d for d in list_one if (d['name'], d['age']) not in have_2]

print(extra)

Results:

[{'name': 'alf', 'age': 50}]

Answer 2

Yet another possible solution:

>>> list(filter(lambda x: not any([set(x.items()).issubset(y.items()) for y in list_two]), list_one))
[{'age': 50, 'name': 'alf'}]

or:

>>> s2 = [set(i.items()) for i in list_two]
>>> list(filter(lambda x: not any([set(x.items()).issubset(y) for y in s2]), list_one))
[{'age': 50, 'name': 'alf'}]

The advantage of this approach is that it does not need to know the "keys" ('age' and 'name') present in both dictionary sets.

Answer 3

Use this:-

new_list = [i for i,j in zip(list_one,list_two) if i['name']!=j['name'] and i['age']!=j['age']]
print (new_list)

Output

[{'name': 'alf', 'age': 50}]

Answer 4

An efficient way would be to convert your two structures to dicts, keyed by the two values, then create the result dict:

key = lambda dct: (dct['name'], dct['age'])
d1 = { key(dct): dct for dct in list_one }
d2 = { key(dct): dct for dct in list_two }
desired_d = { k:v for k,v in d1.items() if k not in d2 }
print(desired_d)
print(desived_d.values())

Answer 5

diff = [
    e for e in list_one
    if (e['name'], e['age']) not in set((e['name'], e['age']) for e in list_two)
]
print diff