CODEWITHSUNDEEP
pythonjsonalgorithmsortingpython-3.6
user19750252
user19750252

Reputation: 21

Transform Python dictionary into List

With Python 3.6, what's the most efficient way to transform this dictionary into a List? I've tried to use a lot of loops, but it doesn't look efficient at all, it takes some time. This is the original dictionary:

d = {'Owner': [{'login': 'AAAA', 'mail': '[email protected]'},
               {'login': 'BBBB', 'mail': '[email protected]'},
               {'login': 'CCCC', 'mail': '[email protected]'}],
     'Stakeholder': [{'login': 'DDDD', 'mail': '[email protected]'},
                     {'login': 'AAAA', 'mail': '[email protected]'}],
     'Team': [{'login': 'CCCC', 'mail': '[email protected]'},
              {'login': 'BBBB', 'mail': '[email protected]'}]}

This is the goal:

[{'login': 'AAAA', 'mail': '[email protected]', 'roles': ['Owner', 'Stakeholder']},
 {'login': 'BBBB', 'mail': '[email protected]', 'roles': ['Owner', 'Team']},
 {'login': 'CCCC', 'mail': '[email protected]', 'roles': ['Owner', 'Team']},
 {'login': 'DDDD', 'mail': '[email protected]', 'roles': ['Stakeholder']}]

Thanks!

Edit 1: So far I could get a list of unique users:

list_users = []

for role, users in old_dict.items():
    for user in users:
        list_users.append(user)

unique_list_of_users = []
for i in range(len(list_users)):
    if list_users[i] not in list_users[i + 1:]:
        unique_list_of_users.append(list_users[i])

for user in unique_list_of_users:
    user["role"] = []

Upvotes: 1

Views: 88

Answers (5)

joaopfg
joaopfg

Reputation: 1287

I believe the most optimal way is to let your dicts that belong to the list of each role hashable. Then, you can make them be the key of another dict that will tell the roles of each login+mail dict. The python code would be:

class Hashabledict(dict):
    def __hash__(self):
        return hash(frozenset(self))


def solve(d):
    roles = dict()

    for key, value in d.items():
        for cur_dict in value:
            hashable_cur_dict = Hashabledict(cur_dict)

            if hashable_cur_dict not in roles:
                roles[hashable_cur_dict] = {key}
            else:
                roles[hashable_cur_dict].add(key)

    answer = list()

    for key, value in roles.items():
        key['roles'] = list(value)
        answer.append(key)

    return answer

An example of use is below:

if __name__ == "__main__":
    d = {'Owner': [{'login': 'AAAA', 'mail': '[email protected]'},
                   {'login': 'BBBB', 'mail': '[email protected]'},
                   {'login': 'CCCC', 'mail': '[email protected]'}],
         'Stakeholder': [{'login': 'DDDD', 'mail': '[email protected]'},
                         {'login': 'AAAA', 'mail': '[email protected]'}],
         'Team': [{'login': 'CCCC', 'mail': '[email protected]'},
                  {'login': 'BBBB', 'mail': '[email protected]'}]}

    test = solve(d)
    print(test)

The output is:

[{'login': 'AAAA', 'mail': '[email protected]', 'roles': ['Owner', 'Stakeholder']}, {'login': 'BBBB', 'mail': '[email protected]', 'roles': ['Owner', 'Team']}, {'login': 'CCCC', 'mail': '[email protected]', 'roles': ['Owner', 'Team']}, {'login': 'DDDD', 'mail': '[email protected]', 'roles': ['Stakeholder']}]

Upvotes: 0

funnydman
funnydman

Reputation: 11346

Or use collections.defaultdict:

import collections
import pprint
 
d = {
    "Owner": [
        {"login": "AAAA", "mail": "[email protected]"},
        {"login": "BBBB", "mail": "[email protected]"},
        {"login": "CCCC", "mail": "[email protected]"},
    ],
    "Stakeholder": [
        {"login": "DDDD", "mail": "[email protected]"},
        {"login": "AAAA", "mail": "[email protected]"},
    ],
    "Team": [
        {"login": "CCCC", "mail": "[email protected]"},
        {"login": "BBBB", "mail": "[email protected]"},
    ],
}

res = collections.defaultdict(dict)
for k, v in d.items():
    for obj in d[k]:
        if not obj['login'] in res:
            res[obj['login']].update(obj)
        res[obj['login']].setdefault('roles', []).append(k)

pprint.pprint(list(res.values()))

Output:

[{'login': 'AAAA', 'mail': '[email protected]', 'roles': ['Owner', 'Stakeholder']},
 {'login': 'BBBB', 'mail': '[email protected]', 'roles': ['Owner', 'Team']},
 {'login': 'CCCC', 'mail': '[email protected]', 'roles': ['Owner', 'Team']},
 {'login': 'DDDD', 'mail': '[email protected]', 'roles': ['Stakeholder']}]

Upvotes: 1

Mortz
Mortz

Reputation: 4879

Here is another solution using itertools -

from itertools import chain, groupby
flattened_items = sorted(chain.from_iterable([list(zip([k]*len(v), v)) for item in d.items() for k, v in (item,)]), key=lambda x: x[1]['login'])

for k, v in groupby(flattened_items, key=lambda x: x[1]['login']):
    all_roles, all_dicts = zip(*v)
    new_dict = {kk: vv for d in reversed(all_dicts) for kk, vv in d.items()}
    new_dict['roles'] = list(all_roles)
    print(new_dict)

What I am doing is to flatten the d.items() into (key, value) pairs, sort it by the values and then aggregate using itertools.groupby

Output

{'login': 'AAAA', 'mail': '[email protected]', 'roles': ['Owner', 'Stakeholder']}
{'login': 'BBBB', 'mail': '[email protected]', 'roles': ['Owner', 'Team']}
{'login': 'CCCC', 'mail': '[email protected]', 'roles': ['Owner', 'Team']}
{'login': 'DDDD', 'mail': '[email protected]', 'roles': ['Stakeholder']}

Upvotes: 0

ljmc
ljmc

Reputation: 5264

Since you have a key in the login and mail field, you can use those to quickly build a dictionnary with the values you need and then access them.

d = {
    "Owner": [
        {"login": "AAAA", "mail": "[email protected]"},
        {"login": "BBBB", "mail": "[email protected]"},
        {"login": "CCCC", "mail": "[email protected]"},
    ],
    "Stakeholder": [
        {"login": "DDDD", "mail": "[email protected]"},
        {"login": "AAAA", "mail": "[email protected]"},
    ],
    "Team": [
        {"login": "CCCC", "mail": "[email protected]"},
        {"login": "BBBB", "mail": "[email protected]"},
    ],
}

tmp = {}

for k, v in d.items():
    for dd in v:
        try:
            tmp[dd["login"]]["roles"].append(k)
        except KeyError:
            tmp[dd["login"]] = dd
            tmp[dd["login"]].update({"roles": [k]})   

list(tmp.values())

gives

[{'login': 'AAAA', 'mail': '[email protected]', 'roles': ['Owner', 'Stakeholder']},
 {'login': 'BBBB', 'mail': '[email protected]', 'roles': ['Owner', 'Team']},
 {'login': 'CCCC', 'mail': '[email protected]', 'roles': ['Owner', 'Team']},
 {'login': 'DDDD', 'mail': '[email protected]', 'roles': ['Stakeholder']}]

Upvotes: 1

The Thonnu
The Thonnu

Reputation: 3624

Try this:

output = []
for k, v in d.items():
    for dct in v:
        for x in output:
            if x['login'] == dct['login']:
                x['roles'].append(k)
                break
        else:
            output.append({**dct, **{'roles': [k]}})
print(output)

Output:

[{'login': 'AAAA', 'mail': '[email protected]', 'roles': ['Owner', 'Stakeholder']},
 {'login': 'BBBB', 'mail': '[email protected]', 'roles': ['Owner', 'Team']},
 {'login': 'CCCC', 'mail': '[email protected]', 'roles': ['Owner', 'Team']},
 {'login': 'DDDD', 'mail': '[email protected]', 'roles': ['Stakeholder']}]

Note: you specified you were on Python 3.6, so {**dct, **{'roles': [k]}} will work for you. It will not work on Python 3.4 or lower. If you are on 3.4 or lower, use:

dct.update({'roles': [k]})
output.append(dct)

Upvotes: 1

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