how to create a list of values of a particular key in python?

Question

The groups_per_user function receives a dictionary, which contains group names with the list of users. Users can belong to multiple groups. Fill in the blanks to return a dictionary with the users as keys and a list of their groups as values.

code

def groups_per_user(group_dictionary):
    user_groups = {}
# Go through group_dictionary
    for ___:
    # Now go through the users in the group
      for ___:
        # Now add the group to the the list of
        # groups for this user, creating the entry
        # in the dictionary if necessary

return(user_groups)

print(groups_per_user({"local": ["admin", "userA"],
    "public":  ["admin", "userB"],
    "administrator": ["admin"] }))

my code

def groups_per_user(group_dictionary):
    user_groups = {}
   # Go through group_dictionary
    for group in group_dictionary:
    # Now go through the users in the group
      for users in group_dictionary[group]:
        group += [group]
        # Now add the group to the the list of
        # groups for this user, creating the entry
        # in the dictionary if necessary

  return(user_groups)

print(groups_per_user({"local":["admin","userA"],"public":["admin","userB"],"administrator": ["admin"]}))

I am not able to figure out how to proceed further than this.

Please format your code properly, it seems the text in bold is part of your code — Shivam Jha
– Shivam Jha, Commented Jul 3, 2020 at 12:33
Your return statement is not at a valid indentation level. — alani
– alani, Commented Jul 3, 2020 at 12:37

Jagpreet · Accepted Answer · 2020-07-03 12:51:47Z

1

Since you've not mentioned exactly what should be the required output. I believe you want unique values to act as keys and their respective key should be clubbed into one value.

For the same, below mentioned code is working.

def groups_per_user(group_dictionary):
    user_groups = {}
    for groups in group_dictionary:
      for user in group_dictionary[groups]:
          if user not in user_groups:
              user_groups[user] = [groups]
          else:
              user_groups[user].append(groups)   

    return(user_groups)


print(groups_per_user({"local": ["admin", "userA"],
    "public":  ["admin", "userB"],
    "administrator": ["admin"] }))

Output:

{'admin': ['local', 'public', 'administrator'], 'userA': ['local'], 'userB': ['public']}

answered Jul 3, 2020 at 12:51

Jagpreet

292 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

alani Over a year ago

It certainly works, but is there any particular reason why you have renamed the variable group used in the question as groups? It only contains one group at a time.

Jagpreet Over a year ago

Every time loop will iterate through one single group. It's a typo, I didn't notice that. group or groups works the same way.

alani · Accepted Answer · 2020-07-03 12:53:27Z

You had made a good start (aside from some issues of formatting code in the question).

You need to append to a list that is a value in user_groups (with key user), first creating it as an empty list if it does not already exist.

def groups_per_user(group_dictionary):
    user_groups = {}
    for group in group_dictionary:
        for user in group_dictionary[group]:
            if user not in user_groups:
                user_groups[user] = []
            user_groups[user].append(group)

    return(user_groups)

print(groups_per_user({"local":["admin","userA"],"public":["admin","userB"],"administrator": ["admin"]}))

In fact, you can use items() to give a sequence of 2-tuples containing (key, value) and unpack those into separate variables containing key and value (here group and users), avoiding the need for indexing using group_dictionary[group]:

def groups_per_user(group_dictionary):
    user_groups = {}
    for group, users in group_dictionary.items():  # <== use of items()
        for user in users:  # <== now we have the value (users) directly
            if user not in user_groups:
                user_groups[user] = []
            user_groups[user].append(group)
    return(user_groups)

borisdonchev · Accepted Answer · 2020-07-03 13:30:03Z

There are already a couple of answers, but I find it worth mentioning that you could also use defaultdict to solve your problem. The usage is pretty straightforward - if the key is not found in the dictionary the defaultdict uses the method provided in the default_factory:

from collections import defaultdict 

def groups_per_user(group_dictionary):

    user_groups = defaultdict(list)

    for group in group_dictionary:
        for user in group_dictionary[group]:
            user_groups[user].append(group)

    return(user_groups)

print(
    groups_per_user({
        "local": ["admin","userA"],
        "public":["admin","userB"],
        "administrator": ["admin"]
    })
)

>>> defaultdict(<class 'list'>, {'admin': ['local', 'public', 'administrator'], 'userA': ['local'], 'userB': ['public']})

Constructing a defaultdict is more expensive than constructing a normal dict, but the performance is better.

Collectives™ on Stack Overflow

how to create a list of values of a particular key in python?

3 Answers 3

2 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related