Transform Python dictionary into List

Question

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"] = []

If you've tried loops, you should add them to your question we SO can see what you've tried. — ljmc
– ljmc, Commented Aug 12, 2022 at 11:20

The Thonnu · Accepted Answer · 2022-08-12 11:27:21Z

1

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)

answered Aug 12, 2022 at 11:27

The Thonnu

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Comments

ljmc · Accepted Answer · 2022-08-12 11:49:52Z

1

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']}]

edited Aug 12, 2022 at 11:49

answered Aug 12, 2022 at 11:33

ljmc

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2 Comments

DarrylG Over a year ago

Your current code doesn't produce what you give as the result (I upvoted before noticing this).

ljmc Over a year ago

@DarrylG good catch, I had not copy/pasted the change for asking permission to asking forgiveness, now I have and it's fixed.

funnydman · Accepted Answer · 2022-08-12 14:17:07Z

1

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']}]

edited Aug 12, 2022 at 14:17

answered Aug 12, 2022 at 13:40

funnydman

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4 Comments

Mortz Over a year ago

One caveat though - this produces final output with the "last" mail key - the OP's expected output wants the "first" mail key

funnydman Over a year ago

@Mortz What do you mean? Not sure I understand

Mortz Over a year ago

For login: AAAA the expected output for mail is mail: [email protected] (5 As) - which comes from Owner - whereas your output is mail: [email protected] (4 As) - which comes from Stakeholder

funnydman Over a year ago

@Mortz Nice catch! I've fixed that

Mortz · Accepted Answer · 2022-08-12 14:11:18Z

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']}

joaopfg · Accepted Answer · 2022-08-12 14:20:02Z

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']}]

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