# identify who the “key connectors” are among data scientists


users = [
    { "id": 0, "name": "Hero" },
    { "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" }
    ]

print("users:")
print(users)

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

print("\nfriendships:")
print(friendships)

# add a list of friends to each user
for user in users:
   print(user)
   print("Adding attr 'friends' to dict entry")
   user["friends"] = []
   print(user, "\n")

print("users:")
print(users)


# populate the lists using the friendships data:
for i, j in friendships:
    # this works because users[i] is the user whose id is i
    users[i]["friends"].append(users[j])      # add i as a friend of j
    users[j]["friends"].append(users[i])      # add j as a friend of i

print("users:")
print(users)



# “what’s the average number of connections?
def number_of_friends(user):
    """how many friends does _user_ have?"""
    return len(user["friends"])             # length of friend_ids list

for user in users:
   print(user, number_of_friends(user) )
   print()
print()

total_connections = sum(number_of_friends(user) for user in users)
print(total_connections)

num_users = len(users)                           # length of the users list
avg_connections = total_connections / num_users  # 2.4

print(avg_connections)

# Find the most connected person (most # of friends)

# we can sort them from “most friends” to “least friends”:

# (1) create a list (user_id, number_of_friends)
num_friends_by_id = [(user["id"], number_of_friends(user))
                      for user in users]

print()
print(num_friends_by_id)
print()

# Python's built-in sorted() function:
#    sorted(list, key=..., reverse=...)


o = sorted(num_friends_by_id,    
           key= lambda x: x[1],   # Use 2nd field as key for sort
           reverse=True)          # largest to smallest

print(o)