Question

Python code

Write a Python program for each of the following: The following is a social network graph. Each edge (link) in the graph represents a friendship between the involved people. The number on the link represents the number of times the involved people communicated with one another. No number on an edge means 0 communication is happening. You are to write a python program that will store the relations between the people involved in this social network somehow, as well as the degree of their communication. You are also going to print the following: A list of all friends of each person in the network (7.5 points). A list of unique friends of friends of each person in the network. (7.5 points) (make reasonable assumptions) The three highest people involved in some kind of direct communication with others in the network. (7.5 points) The list of people who are involved in zero communications with others. (7.5 points) The average number of friends the work. (7.5 points) All person(s) who have more than 4 friends. (7.5 points) 10 points are awarded for a professional evaluation of your submission. Rick Jamie 3 2 4 3 me 9 Bryan 8 Heather Ray O 4 3 2 2 Bas 5 6 7 Derek Patrick Joshua

Answer 1

python program of this question

"network.py'

class network(object):
        """docstring for Network"""
        def __init__(self):
                self.name={
                'me':[['Rick',2],['Jamie',4],['Bryan',0],['Ray',0],['Bas',0],['Heather',8],['Patrick',2],['Derek',0],['Joshua',4]],
                'Rick':[['Bryan',3],['Bas',3],['me',2]],
                'Jamie':[['me',4],['Heather',9],['Bas',0]],
                'Bryan':[['Ray',0],['Rick',3],['Bas',0]],
                'Ray':[['Patrick',5],['Bas',3],['me',0]],
                'Bas':[['me',0],['Ray',3],['Bryan',0],['Rick',3],['Jamie',0],['Heather',0],['Patrick',7]],
                'Heather':[['me',8],['Jamie',9],['Bas',0],['Derek',2],['Joshua',6]],
                'Patrick':[['me',2],['Ray',5],['Bas',7]],
                'Derek':[['me',0],['Heather',2]],
                'Joshua':[['me',4],['Bas',0],['Heather',6]]
                }

        '''' this method print the list of 
                all friend of each person '''
        def list_of_frinend(self):
                print("\n---------------list of all friend of each person -----------------")
                for k,v in self.name.items():
                        print(k + '\t',end=' ')
                        for x in v:
                                print(x[0],end=' ')
                        print();

        ''' unique_friend_of_friend of each persom'''
        def unique_friend_of_friend(self):
                print("\n---------------unique friend of friend of each person-------------")
                for k,v in self.name.items():
                        print(k + '\t:',end=' ')

                        S = set()
                        for x in v:
                                for y in self.name[x[0]]:
                                        S.add(y[0])

                        print(S);


        ''' highest involed people in communication'''
        def highest_involved(self):
                print('\n-------------three highest involved people ------------------')
                
                L = list()
                for k,v in self.name.items():

                        l =list()
                        for x in v:
                                l.append(x[1])

                        L.append([k,max(l)])


                def myFun(a):
                        return a[1]

                L.sort(key=myFun)

                print(L[-3:])


        def list_zero_communication(self):
                print('\n---------------------------list of those people who have zero communication')
                L = list()
                for k,v in self.name.items():

                        l =list()
                        for x in v:
                                l.append(x[1])

                        if min(l) ==0:
                                L.append(k)

                print(L)

        def  average_number_of_friend(self):
                print('\n-------------average number of friend---------------')

                sum=0;
                for v in self.name.values():
                        sum = sum+len(v);

                print(sum//len(self.name))


        def more_than_4_friends(self):
                print('\n-------------name of person who have more than four friend------------------')

                for k,v in self.name.items():
                        if len(v)>4:
                                print(k,end=' ');


x = network();


x.list_of_frinend();
x.unique_friend_of_friend();
x.highest_involved();
x.list_zero_communication();
x.average_number_of_friend();
x.more_than_4_friends();

print("\n\n\nEND Program\n\n")

you can run this program like

python3 newtwork.py

this program output is follow

ravi@ravi-Inspiron-15-3567:-/Desktop/Chegg$ python3 network.py me Rick Jamie Bryan Ray -list of all friend of each person Rick Jamie Bryan Ray Bas Heather Patrick Derek Joshua Bryan Bas me me Heather Bas Ray Rick Bas Patrick Bas me me Ray Bryan Rick Jamie Heather Patrick me Jamie Bas Derek Joshua me Ray Bas me Heather me Bas Heather Bas Heather Patrick Derek Joshua -unique friend of friend of each person- me : {'Jamie', 'Joshua', 'Derek', 'Bryan', 'Ray' 'me 'Heather', 'Patrick', 'Rick', 'Bas'} Rick : {'Jamie', 'Joshua', "Derek', 'Bryan', 'me' 'Ray' "Heather', 'Patrick', 'Rick', 'Bas'} Jamie : {'Jamie', 'Joshua', 'Derek', 'Bryan' 'Ray' 'me "Heather', 'Patrick', 'Rick', 'Bas'} Bryan :{'Jamie', 'Bryan', 'Ray', 'me', 'Heather', 'Patrick', 'Rick', 'Bas'} Ray : {'Jamie', 'Joshua "Derek', 'Bryan', 'Ray', 'me 'Heather' 'Patrick', 'Rick', 'Bas'} Bas : {'Jamie', 'Joshua', 'Derek', 'Bryan', 'Ray' me "Heather', 'Patrick', 'Rick', 'Bas'} Heather : {'Jamie', 'Joshua' "Derek', 'Bryan', 'Ray', "Heather', 'Patrick', 'Rick', 'Bas'} Patrick : {'Jamie', 'Joshua', 'Derek', 'Bryan', 'Ray', 'me' "Heather', 'Patrick', 'Rick'. 'Bas'} Derek :{'Jamie', 'Joshua', 'Derek', 'Bryan', 'Ray 'me 'Heather', 'Patrick', 'Rick', 'Bas'} Joshua : {'Jamie', 'Joshua', 'Derek', 'Bryan', 'Ray 'me "Heather', 'Patrick', 'Rick', 'Bas'} 'me' three highest involved people [['me', 8], ['Jamie', 9], ['Heather', 9]] -list of those people who have zero communication ['me', 'Jamie', 'Bryan', 'Ray', 'Bas', 'Heather', 'Derek', 'Joshua'] -- avarage number of friend - 4 --name of person who have more than four friend - me Bas Heather END Program ravi@ravi-Inspiron-15-3567:-/Desktop/Cheggs |