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Find mutual friends count from given relation ship array. Substantive Expertise

This is a good example of the “substantive expertise” aspect of data science.

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" }
]
friendship_pairs = [(0, 1), (0, 2), (1, 2), (1, 3), (2, 3), (3, 4),
                    (4, 5), (5, 6), (5, 7), (6, 8), (7, 8), (8, 9)]
# Initialize the dict with an empty list for each user id:
friendships = {user["id"]: [] for user in users}
# And loop over the friendship pairs to populate it:
for i, j in friendship_pairs:
    friendships[i].append(j)  # Add j as a friend of user i
    friendships[j].append(i)  # Add i as a friend of user j
 
def number_of_friends(user):
    """How many friends does _user_ have?"""
    user_id = user["id"]
    friend_ids = friendships[user_id]
    return len(friend_ids)

total_connections = sum(number_of_friends(user) for user in users)        # 24
num_users = len(users)                            # length of the users list
avg_connections = total_connections / num_users   # 24 / 10 == 2.4
# Create a list (user_id, number_of_friends).
num_friends_by_id = [(user["id"], number_of_friends(user)) for user in users]
num_friends_by_id.sort(key=lambda id_and_friends: id_and_friends[1],reverse=True)# Sort the list,by num_friends,largest to smallest
def foaf_ids_bad(user):
    """foaf is short for "friend of a friend" """
    val=[]
    friend_ids=[]
    for friend_id in friendships[user["id"]]:
        for foaf_id in friendships[friend_id]:
            val.append(foaf_id)
    return val                   
from collections import Counter 
def friends_of_friends(user):
    user_id = user["id"]
    return Counter(
        foaf_id
        for friend_id in friendships[user_id]     # For each of my friends,
        for foaf_id in friendships[friend_id]     # find their friends
        if foaf_id != user_id                     # who aren't me
        and foaf_id not in friendships[user_id]   # and aren't my friends.
    )
print(friends_of_friends(users[3])) 
print(friendships[users[0]['id']])
print(foaf_ids_bad(users[3]))
print(num_friends_by_id)
print(avg_connections)
print(num_users)
print(total_connections)
print(friendships)
print(friendship_pairs)
print(users[0]['id'])

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Books To Refer

1. C#   in Depth ,  Third   Edition : Foreword   by Eric   Lippert        By:  Jon Skeet    Publication Date:  16-SEP-2013 2. Learning jQuery - Fourth Edition     B y:  Jonathan Chaffer; Karl Swedberg     Pub. Date:  June 25, 2013      3.Beginning JSON                    By:  Ben Smith     Publication Date:  04-MAR-2014 4. Responsive Web Design with jQuery     By:  Gilberto Crespo      Pub. Date:  November 25, 2013 5. Developing Responsive Web Applications with AJAX and jQuery       By:  Sandeep Kumar Patel  P ub. Date:  July 25, 2014 6. Dependency Injection in .NET     by  Mark Seemann