1. Sum of values of Dictionary Given a dictionary, which both key and value are integer. return the sum of all values which the key is > n Input: {10: 1, 20: 3, 30: 4, 40:5}, n = 15 Output: 12 Input: {10: 1, 20: 3, 30: 4, 40: 5}, n = 25 Output: 9 Input: {10: 1, 20: 3, 30: 4, 40: 5}, n = 35 Output: 5 10 11 12 13 14 15 16 17 89972 18 19 20 21 22 # # Complete the 'sum_of_dict_value' function below. # # The function is expected to return an INTEGER. # The function accepts following parameters: # 1. Dict dictionary # 2. INTEGER n # def sum_of_dict_value(dictionary, n): # Write your code here

1. Sum of values of Dictionary Given a dictionary, which both key and value are integer. return the sum of all values which the key is > n Input: {10: 1, 20: 3, 30: 4, 40:5}, n = 15 Output: 12 Input: {10: 1, 20: 3, 30: 4, 40: 5}, n = 25 Output: 9 Input: {10: 1, 20: 3, 30: 4, 40: 5}, n = 35 Output: 5 10 11 12 13 14 15 16 17 89972 18 19 20 21 22 # # Complete the 'sum_of_dict_value' function below. # # The function is expected to return an INTEGER. # The function accepts following parameters: # 1. Dict dictionary # 2. INTEGER n # def sum_of_dict_value(dictionary, n): # Write your code here

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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A date-to-event dictionary is a dictionary in which the keys are dates (as strings in the format 'YYYY-MM-DD'), and the values are lists of events that take place on that day. Complete the following function according to its docstring. Your code must not mutate the parameter! def group_by_year (events: Dict[str, List[str]]) -> Dict [str, Dict [str, List[str]]]: """Return a dictionary in which in the keys are the years that appear in the dates in events, and the values are the date-to-event dictionaries for all dates in that year. >>> date_to_event = {'2018-01-17': ['meeting', 'lunch'], '2018-03-15': ['lecture'], '2017-05-04': ['gym', 'dinner']} >>> group_by_year (date_to_event) {'2018': {'2018-01-17': ['meeting', 'lunch'], '2018-03-15': ['lecture']}, '2017': {'2017-05-04': ['gym', 'dinner']} } || || ||
def max_kv(d): """ Given a dictionary, return a tuple of key value pair where the key is the largest in d. """
Write a program that creates a dictionary containing course numbers and the room numbers ofthe rooms where the courses meet. The dictionary should have the following key-value pairs:Course Number (key) Room Number (value)CS101 3004CS102 4501CS103 6755NT110 1244CM241 1411The program should also create a dictionary containing course numbers and the names of theinstructors that teach each course. The dictionary should have the following key-value pairs:Course Number (key) Instructor (value)CS101 HaynesCS102 AlvaradoCS103 RichNT110 BurkeCM241 LeeThe program should also create a dictionary containing course numbers and the meeting timesof each course. The dictionary should have the following key-value pairs:Course Number (key) Meeting Time (value)CS101 8:00 a.m.CS102 9:00 a.m.CS103 10:00 a.m.NT110 11:00 a.m.CM241 1:00 p.m.The program should let the user enter a course number, then it should display the course’s roomnumber, instructor, and meeting time.
Read strings from input until 'stop' is read. For each string read: If the string is a key in dictionary patients_record, delete the key from patients_record and output the string followed by ' deleted'. Otherwise, output the string followed by ' not a key'.
def invert_dict(d): """ Given a dictionary d, create the invert of the dictionary where the key is a value v in d, and v is mapped to a set of keys in original d whose values are equal to v. Return the inverted dictionary mapping value to set of keys """
Function in python thst inverts the key and value in a dictionary If dictionary encounters more than one of the same key when trying to invert dictionary. Raise a key error
Create a dictionary (order_dict) that the key is the burger number, and the value is a list of the quantities. For example: order_dict which is {'1': [10, 3, 78], '2': [35, 0, 65], '3': [9, 23, 0], '4': [0, 19, 43], '5': [43, 0, 21]} Create a dictionary (total_order_dict) that the key is the burger number, and the value is the total of the quantities for that burger. For example: total_order_dict which is {'1': 91, '2': 100, '3': 32, '4': 62, '5': 64} Use "assert to test the values of the total_order_dict For example: expected_result_ototal_order_dict = {'1': 91, '2': 100, '3': 32, '4': 62, '5': 64} assert actual_result_ototal_order_dict['1'] == expected_result_ototal_order_dict['1'] , "The actual result is not the \same as expected result for the order number 1!" (and test the other elements) The function can be like: def test_sum(): #Testing assert actual_result_ototal_order_dict['1'] == 91, "The actual result is not the \ same as expected result for the…
get_top_regions() that takes three input parameters; the 2-D list (similar to the database), the dictionary that stores information of all regions, and a non-zero positive integer x. This function returns a new dictionary that contains the information of top-x regions in terms of the total number of hospitalization cases recorded in these regions. That means these top-x regions have the largest numbers of hospitalization cases. Each {key:value} pair in this resulting dictionary stores the name of a top region as the key and the value is the total number of hospitalization cases reported in this region. The result does not need to be sorted, the regions of the resulting dictionary can appear in any order as long as the top-x regions have been identified correctly. If the value of x is less than 1, or more than the total number of unique regions, your function should print a message informing the user and return an empty dictionary. See sample outputs below. >>> topx_regions…
Implement the function merge_dict. The merge_dict function merges two dictionaries with the same keys together by adding up their values for the corresponding keys and returning the resulting dictionary. def merge_dict(d1, d2): """Returns a dictionary that is the result of two dictionaries being merged together. Dictionaries are merged by adding up their values. You can assume that the same keys appear in both dictionaries. >>> data8 = {"midterms":1, "projects":3} >>>data100 = {"midterms":2, "projects":3} >>> combined_exams = merge_dict (data8, data100) >>> combined_exams {'midterms': 3, 'projects': 6} >>> sunday_orders = {"pizza": 3, "hot dogs": 2, "fries": 5} >>> monday_orders = {"pizza": 1, "hot dogs": 1, "fries": 8} >>> combined_orders = merge_dict (sunday_orders, monday_orders) >>> combined_orders {'pizza': 4, 'hot dogs': 3, 'fries': 13} "*** YOUR CODE HERE ***'
True/False 9. A Python dictionary is a kind of sequence.
Please correct the code for this lab. Current code: import random def create_random_open_ports(): #Create empty list / dictionary port = [] ports = {} #Fill list with 100 random port numbers from 10 - 100 for i in range(1,101): random_port = random.randint(10, 100) port.append(random_port) #Create a new dictionary with ports numbers as keys new_ports = ports.fromkeys(port, 0) #Iterate through new dictionary add random 0 / 1 #Closed port = 0, Open port = 1 for key in new_ports: r1 = random.randint(0,1) new_ports[key] = r1 return new_ports # TODO: create a function called create_host_IPs() that randomly generates a list of 10 IP addresses # concatenating the 4 values into a string and appending it to a host list. The function returns the generated # host list def create_host_IPs(): host = [] octet1 = "" octet2 = "" octet3 = "" octet4 = "" #Add loop with range 1 - 11 to create the 10 host IPs EX: 192.123.11.1…
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