• 7.Given this nest dictionary grab the word "MDS":"outer": [1,2,3,{"inner": ["this", "is", "inception", {"inner_inner": [1, 2, 3, "MDS"]}]),# Your answer here.• 8.Why does the following cell return an error?(] t= (1, 2, 3, 4, 5)t[-1] - 6TypeError in 1t = (1, 2, 3, 4, 5)----> 2 t[-1] = 6Traceback (most recent call last)TypeError: 'tuple object does not support iten assignnentSEARCH STACK OVERFLOW

List indexing starts from 0 and the elements in list can be extracted using indexing and the value…

Given this nest dictionary grab the word "MDS": "outer": I 1, 2, 3, {"inner": ["this", "is", "inception", {"inner_inner": [1, 2, 3, "MDS"]}]}, I Your answer here. 8. Why does the following cell return an error? U t- (1, 2, 3, 4, 5) t(-1] - 6

Given this nest dictionary grab the word "MDS": "outer": I 1, 2, 3, {"inner": ["this", "is", "inception", {"inner_inner": [1, 2, 3, "MDS"]}]}, I Your answer here. 8. Why does the following cell return an error? U t- (1, 2, 3, 4, 5) t(-1] - 6

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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java question 32. need help thank u
Here is my entire code(average is not working yet). The response I got from the professor is that I have to create a DRY solution replacing the five display_functions with a single get_array function you call five times like common_array = get_array(plant_xml, "COMMON") etc. from urllib.request import urlopenfrom xml.etree.ElementTree import parse def read_data(data_file): plant_url = urlopen(data_file) plant_xml = parse(plant_url) return plant_xml def display_common_items(plant_xml): plant_root = plant_xml.getroot() common_array = [] for item in plant_root.findall("PLANT"): common_array.append(item.find("COMMON").text) return common_array def display_botanical_items(plant_xml): plant_root = plant_xml.getroot() botanical_array = [] for item in plant_root.findall("PLANT"): botanical_array.append(item.find("BOTANICAL").text) return botanical_array def display_zone_items(plant_xml): plant_root = plant_xml.getroot() zone_array = []…
Analyze the given linked list figure, use the following pointers (head, p1, p2, p3, p4) and choose the correct answer: head 20 30 10 p1 data p2 tmp next p3 20 p4 NULL . O p4->next is NULL p4 is NULL O p4 is equal to p3 O p4->data is 5
Write a program to convert the following two lists into the wanted dictionary keys ['Ten', 'Twenty. Thirty values [10, 20, 30] The wanted dictionary: (Ten': 10, 'Twenty': 20, 'Thirty': 30).
Given the following Java functions, transcribe it into a Python 3 program. /*** Add a new entry with the specified key, value and hash code to* the specified bucket. It is the responsibility of this* method to resize the table if appropriate.*/void addEntry(int hash, Object key, Object value, int bucketIndex) {table[bucketIndex] = new Entry(hash, key, value, table[bucketIndex]);if (size++ >= threshold)resize(2 * table.length);}
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 """
Task 2. Instead of using the default hashCode method to each key, create a hash function applying the folding method. Within your hash function Split the string into 3 sub-strings, based from the reg-ex "-" Parse your sub-strings to Integer and add (folding method) Parse the summed value to String Initialize an int variable for your hash to 0 Loop to each character index of the String Apply the general formula for hashing: hash=31*hash+character in that index position; Append & 0xFFFFFFF to it to make the resulting hash positive Convert the hash to index of tablesize=20 using the % operator Return hash What index has the most occurrence of collisions?
Using hash tables of a Map data structure you will write. Write your application by following the steps below: 1. Store char * (name) as key and int (number of courses taken) as value in your map. Let the hash function you use use an algorithm you specify when converting the name into an index. 2. Let the put function add the given name and the number of lessons taken to the map. If the name (key) exists on the map, replace this existing name and the number of courses taken with a new one. That is, only one unique key can be found on the map at a time. 3. The contains function checks whether the key (name) sent to it is on the map. Send 1 if available, 0 otherwise. 4. Let the Get function send the number of lessons learned for the most sent key. If the key is not present in the map, it returns -1. 5. The display function displays all key-value pairs of the map print it. 6. Search for free space linearly in case of collision. Solve the question in C language?
Need answer within 30 minutes. Provide outuput also
8. names_with_addresses It's time to mail out our form letters from homework 4! Given a friends database of the following shape (a list of dictionaries), return a list of tuples of friends names with their addresses, in the same order they are given in the input list. friends_database = [ {"name":"bimmy", "height":600, "address" :"in the woods"}, {"name":"Dweezil", "height": 175, "address":"123 Main Street"}, {"name":"Emily", "height": 190, "address":"7070 Willow Lane"} Keep in mind that your function should work with any friends database, not just this example one! It's passed as a parameter! Sample calls should look like this: >> names_with_addresses (friends_database) [('bimmy', 'in the woods'), ('Dweezil', '123 Main Street'), ('Emily', '7070 Willow Lane')] >>> names_with_addresses ([]) []
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() %3D >>> db.add_ friend(friend1) >>> db.add_friend(friend2) >>> b.add_friend (friend3) >>> db.group_friends_by_food() {'strawberries': ['bimmy', 'sarah'], 'pizza': ['bimmy', 'dweezil']}
This is a data structure & algorthim question write correctly and do not plagarise please