7. group_friends_by_food For this problem, you are given a mostly-working version of the Friend and FriendsDB classes from hw6 and hw7, and we will add new method to FriendsDB -- group_friends_by_food. group_friends_by_food returns a dictionary mapping from each of the favorite foods enjoyed by any friend to a list of the friends who enjoy that food, sorted in alphabetical order. (You can use the friends_who_love method to generate these lists.) A sample run should look like this: >>> friend1 - Friend("sarah", 165) >>> friend1.add favorite food("strawberries") >>> friend2 Friend("dweezil" 175) >>> friend2.add favorite food ("pizza") >>> friend3- Friend ("bimmy" 60) >>> friend3.add favorite food("pizza") >>> friend3.add favorite food ("strawberries"). >>> db = FriendsDB() >>> db.add friend(friend1) >>> db.add friend(friend2) >>> db.add friend(friend3) >>> db.group_friends_by_ food() {'strawberries'; ['bimmy', 'sarah'], 'pizza': ['bimmy', 'dweezil']}

def get_nearest_station(my_latitude: float, my_longitude: float, stations: List['Station']) -> int: "''Return the id of the station from stations that is nearest to the location given by my_latidute and my_longitude. In the case of a tie, return the ID of the last station in stations with that distance. Preconditions: len(stations) > 1 » get_nearest_station(43.671134, -79.325164, SAMPLE_STATIONS) 7571 » get_nearest_station(43.674312, -79.299221, SAMPLE_STATIONS) 7486

Numpy

def upgrade_stations(threshold: int, num_bikes: int,                     stations: List["Station"]) -> int:    """Modify each station in stations that has a capacity that is less than    threshold by adding num_bikes to the capacity and bikes available counts.    Modify each station at most once.     Return the total number of bikes that were added to the bike share network.     Precondition: num_bikes >= 0

Assume, you have been given a dictionary where the keys present the name of the father and the value is a list of names of the sons of that person. For example A has 3 sons namely X, Y, and Z. Again X has three sons namely E, F, and G. So, A is the grandfather of E, F, and G. family = {"A" : ["X", "Y", "Z"], "B": ["M", "N"], "W" : ["A", "B"], "X" : ["E", "F", "G"]} You need to write a code which takes an input and prints the names of all his grandsons in the format shown in output samples. If he does not have any grandchildren just print “Get your sons married first! Wanna_be_grandpa!!” You can assume the input will always be a key from the dictionary. But we will check your code with a different dictionary. So do not write a code for this particular dictionary only.                 Enter                       Assume, you have been given a dictionary where the keys present the name of the father and…

7. group_friends_by_food For this problem, you are given a mostly-working version of the Friend and FriendsDB classes from hw6 and hw7, and we will add new method to FriendsDB -- group_friends_by_food. group_friends_by_food returns a dictionary mapping from each of the favorite foods enjoyed by any friend to a list of the friends who enjoy that food, sorted in alphabetical order. (You can use the friends_who_love method to generate these lists.) A sample run should look like this: |>>> friend1 = Friend ("sarah", 165) >>> friend1.add_favorite_food ("strawberries") >>> friend2 - Friend ("dweezil", 175) >>> friend2.add_favorite_food("pizza") >>> friend3 = Friend("bimmy", 60) >>> friend3.add_favorite_food("pizza") >>> friend3.add_favorite_food("strawberries") >>> db = FriendsDB() >>> db.add friend (friend1) >>> db.add_friend(friend2) >>> db.add friend(friend3) >>> db.group_friends_by_food() {'strawberries': ['bimmy', 'sarah'], 'pizza': [' bimmy', 'dweezil']}

Is there a difference between the void and the NULL pointers? Your response should provide relevant instances to support it. Distinguish between the two in TABULAR format.

create a class Person to represent a person and another one to represent the database. A dictionary should be a data member of the database class, with Person objects as search keys for this dictionary. Write a program that tests and demonstrates your database.

given dictionaries, d1 and d2, create a new dictionary with the following property. For each entry(a,b) in d1, if a is not a key of d2 then add (a,b) to the new dictionary. For each entry (a,b) in d2, if a is not a key of d1 then add (a,b) to the new dictionary. For example, if d1 is {2:3, 8:19, 6:4, 5:12} and d2 is {2:5, 4:3, 3:9}, then the new dictionary should be {8:19, 6,4, 5:12, 4:3, 3:9}. using puthon

def get_nearest_station(my_latitude: float, my_longitude: float, stations: List['Station']) -> int:    """Return the id of the station from stations that is nearest to the    location given by my_latidute and my_longitude.     In the case of a tie, return the ID of the last station in stations with    that distance.     Preconditions: len(stations) > 1     >>> get_nearest_station(43.671134, -79.325164, SAMPLE_STATIONS)    7571    >>> get_nearest_station(43.674312, -79.299221, SAMPLE_STATIONS)    7486    """