Python Dict

Sample code for working with Python Dicts

# init users list of dicts, print out id 1, then init friends list of tuples

users = [

     { "id": 0, "name": "Bob" },
     { "id": 1, "name": "Dunn" },
     { "id": 2, "name": "Sue" },
     { "id": 3, "name": "Chi" },
     { "id": 4, "name": "Thor" },
     { "id": 5, "name": "Clive" },
     { "id": 6, "name": "Hicks" },
     { "id": 7, "name": "Devin" },
     { "id": 8, "name": "Kate" },
     { "id": 9, "name": "Klein" },
     { "id": 10, "name": "Jen" } 

]

i=0
for user in users:
     if(users[i]["name"]=="Bob"):
         print("Bob ID: ",users[i]["id"])
     i+=1

friends= [(0, 1), (0, 2), (1, 2), (1, 3), (2, 3), (3, 4), (4, 5), (5, 6), (5, 7), (6, 8), (7, 8), (8, 9)]

# spin through all users and create empty list to store list, then populate

for user in users:
     user["friends"]=[]

# populate empty list with all left side of tuple with right and vice versa

for i,j in friends:
     users[i]["friends"].append(users[j])
     users[j]["friends"].append(users[i])

# function to return length based on passed in user

def number_of_friends(user):     
    """how many friends does _user_ have?"""
    return len(user["friends"]) # length of friend_ids list

# total up all friends

total_connection = sum(number_of_friends(user)                         
     for user in users) # 24 

# grab number of users

num_users = len(users)

avg_connections = total_connection / num_users # 2.4 

# create a list (user_id, number_of_friends)

num_friends_by_id = [(user["id"], number_of_friends(user)) for user in users] 

print(sorted(num_friends_by_id,key=lambda pair: pair[1],                      reverse=True))

# Output – largest to smallest

[(1, 3), (2, 3), (3, 3), (5, 3), (8, 3), (0, 2), (4, 2), (6, 2), (7, 2), (9, 1), (10, 0)]

If you have found my website useful, please consider buying me a coffee below 😉