
Database System Concepts
7th Edition
ISBN: 9780078022159
Author: Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher: McGraw-Hill Education
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- 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 = []…arrow_forwardIn c++ i have this information Type of Ticket &. Cost Monday. Adult= £25 and Child = £15 Thursday. Adult = £40 and child = £20 I have made a vector to store this information Vectorticket ={“Monday”, “Thursday”}; Vectoradult = {25, 40}; Vectorchildren ={15, 20}; Thr format of the adult and children vectors is in the order such that the first elements in those vectors correspond with the first element in the ticket vector Now what am struggling with, is i want the program to calculate the total cost of the tickets in correspondant with The user’s inputs which will be the quantity of how many tickets the user bought respectively (adult and children) So something like Monday 1 2 So 1 will be quantity of adult ticket and 2 will quantity of children ticket and the total costs of both added togetherarrow_forwardMATLAB: Create a vector named n which starts at -10 and steps by 1 to end at +10 by using the start:step:stop syntax. Next, generate a vector x1 so that x1 satisfies the mathematical relationship ?ଵ[?]=(0.60). Since n is a vector, one will need to utilize 0.60.^n (i.e., .^ is used in place of ^) to define x1. Redo the work just completed, but for a vector x2 where x2 satisfies the mathematical relationship ?ଶ[?]=(0.85). Repeat the work just completed, but for a vector x3 where x3 satisfies the mathematical relationship ?ଷ[?]=(1.10). In one figure window, plot x1 as a function of n using the stem command. The stem plot must show BLACK (versus the default blue) lines that have filled circles and a line width of 2; refer to the lecture example script file. Further, the horizontal axis label (xlabel) must be n, the title (title) must include your name and the vertical axis label (ylabel) must show ?ଵ[?]=(0.60)(note the subscript and superscript). In addition, turn on the grid (doc grid),…arrow_forward
- Determine if the following statement are true or false screen shot shows the text's SortedArrayCollection and ArrayCollection When comparing two objects using the == operator, what is actually compared is the references to the objects. When comparing two objects using the equals method inherited from the Object class, what is actually compared is the references to the objects. The text's sorted array-based collection implementation stores elements in the lowest possible indices of the array. The text's array-based collection implementation stores elements in the lowest possible indices of the array.arrow_forwardIntegers are read from input and stored into a vector until 0 is read. If the vector's last element is odd, output the odd elements in the vector. Otherwise, output the even elements in the vector. End each number with a newline. Ex: If the input is -9 12 -6 1 0, the vector's last element is 1. Thus, the output is: -9 1 Note: (x % 2!= 0) returns true if x is odd. 4 5 int main() { 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20} vector vect1; int value; int i; bool isodd; cin >> value; while (value != 0) { vect1.push_back(value); cin >> value; } V* Your code goes here */ return 0; 2 3arrow_forwardAn_array and a number kFind the max elements of each of its sub-arrays of length k. Keep indexes of good candidates in deque d.The indexes in d are from the current window, they're increasing,and their corresponding nums are decreasing.Then the first deque element is the index of the largest window value. For each index i: 1. Pop (from the end) indexes of smaller elements (they'll be useless).2. Append the current index.3. Pop (from the front) the index i - k, if it's still in the deque (it falls out of the window).4. If our window has reached size k, append the current window maximum to the output..arrow_forward
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