= 0 for value in y: if len(value) > x:
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Enter a list of at least three elements for y such that the value of x is 2 when the loop completes.
x = 0
for value in y:
if len(value) > x:
x = len(value)
Step by step
Solved in 2 steps
- a and b are int variables containing values, and a < b.Write a for loop that prints all odd numbers from a to b (both inclusive).(Example: if a=25, b=82, then it will print 25 27 29 .... 79 81)Write only the for loop with its block, nothing else. with javaTrue or False: The initial loop is more efficient than its vecotized equivalent?Write a for loop to print all elements in courseGrades, following each element with a space (including the last). Print forwards, then backwards. End each loop with a newline. Ex: If courseGrades = {7, 9, 11, 10}, print:7 9 11 10 10 11 9 7 #include <iostream>using namespace std; int main() { const int NUM_VALS = 4; int courseGrades[NUM_VALS]; int i; for (i = 0; i < NUM_VALS; ++i) { cin >> courseGrades[i]; } /* Your solution goes here */ return 0;}
- Java Task Write a for loop that prints the characters stored in a String object backward. The String is entered by the user. ***Use ch = phrase.charAt(i); to get the character that is in index i of the String phrase.**** For example The user enters “Good morning!” The printout should be “!gninrom dooG”.Code in R Write a while loop what prints all numbers up to 20, but it skips a collection of numbers: 3,9,13,19. A next statement is useful when we want to skip the current iteration of a loop without terminatingit. On encountering next, the R parser skips further evaluation and starts next iteration of the loop. x <- 1:5for (val in x) {if (val == 3){next}print(val)}Q3. Write for-loops and while-loops to:(a) Compute the mean of every column in mtcars.(b) Create an object to store unique values in each column of iris. (Hint: Use function unique()) Theobject could be whatever object you like, list, matrix, dataframe, etc.
- Variable t has been assigned a tuple whose elements are numbers. Write a code that uses while loop that counts the number of times the first element of the tuple appears in the rest of the tuple and assign the number to a variable named repeats. Code: the wrong message Solutions with this approach don't use comma t=(1,6,5,7,1,3,4,1) repeats=0 i=1 while i<len(t): if t[i]==t[0]: repeats+=1 i+=1Consider the following code segment. What is the value of i and sum2 after the for loop completes? int sum2 = 0, i = 0; for (int i = 0; i < 5; i++) sum2 = sum2 + i; cout << "Sum is: " << sum2 << " i is: " << i << endl;Write a for loop to print all NUM_VALS elements of vector courseGrades, following each with a space (including the last). Print forwards, then backwards. End with newline. Ex: If courseGrades = {7, 9, 11, 10}, print:7 9 11 10 10 11 9 7 Hint: Use two for loops. Second loop starts with i = courseGrades.size() - 1 (Notes)C++ without using std:: Note: These activities may test code with different test values. This activity will perform two tests, both with a 4-element vector (vector<int> courseGrades(4)). See "How to Use zyBooks".Also note: If the submitted code has errors, the test may generate strange results. Or the test may crash and report "Program end never reached", in which case the system doesn't print the test case that caused the reported message. #include <iostream>#include <vector>using namespace std; int main() { const int NUM_VALS = 4; vector<int> courseGrades(NUM_VALS); int i; for (i = 0; i < courseGrades.size(); ++i) { cin >>…
- Create a Flowchart for this code. #Declaring and initializing the listcars = ["Buick","Ford","Honda"]colors = ["Red","Blue","Black","White","Green"]aType = ["Sedan","SUV","Sports","Coupe","Truck"]#Using list comprehension to generate a listarr = [[0 for i in range(len(cars))] for j in range(len(cars)*len(colors))]#Outer for loopfor i in range(0, len(colors)*len(cars)): #Inner for loop for j in range(0, len(cars)): #Check if j =0 if j == 0: #Initializing the list with cars(i%5) arr[i][j] = cars[int(i/5)] #chck if j = 1 if j == 1: #Initializing the list with colors(i%5) arr[i][j] = colors[int(i % 5)] #check if j = 2 if j == 2: #Initializing the list with aType(i%5) arr[i][j] = aType[int(i % 5)]#Iterating through the 2D listfor i in range(0, len(arr)): #Print each list one by one print(arr[i])Transcribed Image Text Implement the following to search a letter in a list. a) Create a list named vehicles: car, Truck, boat, PLANE. b) Request a user input for a search letter. c) Use the decision structure in a for loop to search all the items which contains the input letter (ignoring case) in the list. d) Print the item and its position in vehicles if it exists. Otherwise, print the statement indicating it does not contain the letter to search. e) Print the error message if more than one letter is entered. Example Output 1 Vehicles = i'car', 'Truck', 'boat', 'PLANE’] Enter a search letter: a car contains 'a' and it is in position 0. Truck does not contain 'a'. boat contains 'a' and it is in position 2. PLANE contains 'a' and it is in position 3. Example Output 2 Vehicles = ['car', 'Truck', 'boat', 'PLANE'] Enter a search letter: A car contains 'A' and it is in position 0. Truck does not contain 'A'. boat contains 'A' and it is in position 2. PLANE contains 'A' and it is in…A for loop can contain multiple initialization actions separated by commas. True False in java
