I have 2 lists, I want to merge them, answer should be like below

user_16503418 0
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I have two lists of dictionaries and I'd like to merge them. When a dictionary is present in both lists, I'd like to add a "confidence" key to the dictionary to reflect that the dictionary was present in both lists.

List-1

lst1 = [
    {'key': 'data_collected.service_data'},
    {'key': 'gdpr.gdpr_compliance'},
    {'key': 'disclosure_of_information.purpose_of_disclosure'},
    {'key': 'opt_out.choice_of_opt_out'}
]

List-2

lst2 = [
    {'key': 'child_data_protection.parent_guardian_consent'},
    {'key': 'ccpa.ccpa_compliance'},
    {'key': 'disclosure_of_information.purpose_of_disclosure'},
    {'key': 'opt_out.choice_of_opt_out'}
]

when i run below code i am not getting proper output

res = []
for x in lst1:
    for y in lst2:
        if x["key"] == y["key"]:
            if x not in res and y not in res:
                res.append({"key": x["key"], "confidence": 1})
        else:
            if x not in res and y not in res:
                res.append(x)
                res.append(y)

print(res)

OUTPUT should like

[
    {'key': 'data_collected.service_data'},
    {'key': 'gdpr.gdpr_compliance'},
    {
        'key': 'disclosure_of_information.purpose_of_disclosure',
        'confidence': 1
    },
    {
        'key': 'opt_out.choice_of_opt_out',
        'confidence': 1
    },
    {'key': 'child_data_protection.parent_guardian_consent'},
    {'key': 'ccpa.ccpa_compliance'}
]
concatenation python-3.x list dictionary python
5 Answers

lst1.extend(i for i in (i if i not in lst1 else lst1[lst1.index(i)].update({'confidence': 1}) for i in lst2) if i is not None)

lst1 will be your result

user_117570010

You can use set comprehension to collect the "key" elements of each dictionary in your list. Then you can loop through all keys and check whether a key is in both lists.

keys_1 = {d["key"] for d in lst1}
keys_2 = {d["key"] for d in lst2}

output = []
for k in keys_1 | keys_2:
    d = {"key": k}
    if k in keys_1 and k in keys_2:
        d["confidence"] = 1
    output.append(d)
user_46220650

You can avoid raw loops entirely using the intersection and symmetric_difference function on set:

# Shortened key names for brevity
a = [{"key": "a"}, {"key": "b"}, {"key": "c"}]
b = [{"key": "a"}, {"key": "d"}, {"key": "e"}]

# Turn both lists into sets
a_keys = {entry["key"] for entry in a}
b_keys = {entry["key"] for entry in b}

# Add elements that are in both sets with confidence set to 1
result = [{"key": key, "confidence": 1} for key in a_keys.intersection(b_keys)]
# Add elements that are not in both sets
result += [{"key": key} for key in a_keys.symmetric_difference(b_keys)]

Will result in:

[{'confidence': 1, 'key': 'a'},
 {'key': 'b'},
 {'key': 'd'},
 {'key': 'c'},
 {'key': 'e'}]

Note, that the element order will change, as they went through a set.

user_69247400

If you are not too worried about performance.

intersection = [value for value in lst1 if value in lst2]
res = [val for val in lst1 if val not in intersection] + [val for val in lst2 if val not in intersection]
res += list(map(add_confidence, intersection))
user_149331940

Another approach can be:

lst1 = [
    {'key': 'data_collected.service_data'},
    {'key': 'gdpr.gdpr_compliance'},
    {'key': 'disclosure_of_information.purpose_of_disclosure'},
    {'key': 'opt_out.choice_of_opt_out'}
]
lst2 = [
    {'key': 'child_data_protection.parent_guardian_consent'},
    {'key': 'ccpa.ccpa_compliance'},
    {'key': 'disclosure_of_information.purpose_of_disclosure'},
    {'key': 'opt_out.choice_of_opt_out'}
]
for data in lst1:
    # If same data exists in lst2, add confidence key and remove it from lst2
    if data in lst2:
        lst2.remove(data)
        data['confidence']=1

# At the end of above for loop, lst2 contains unique data, now just add both the lists to get the final result            
lst1 = lst1+lst2        
print (lst1)

Output:

[{'key': 'data_collected.service_data'}, {'key': 'gdpr.gdpr_compliance'}, {'key': 'disclosure_of_information.purpose_of_disclosure', 'confidence': 1}, {'key': 'opt_out.choice_of_opt_out', 'confidence': 1}, {'key': 'child_data_protection.parent_guardian_consent'}, {'key': 'ccpa.ccpa_compliance'}]
user_39495640
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