write a python program to find the" kth smallest element" in the ascending order sorted list. also check if the given number is present in the list using any searching technique is suppose element to be searched is not available then display "not present" otherwise display" present"

The chapter is nested lists Create code in python using for loops Thanks!

Generate a random list of 150 integers. Calculate the Standard Deviation of the List. Ask the user for an integer and determine if its exceptional or not.

python LAB: Subtracting list elements from max When analyzing data sets, such as data for human heights or for human weights, a common step is to adjust the data. This can be done by normalizing to values between 0 and 1, or throwing away outliers. Write a program that adjusts a list of values by subtracting each value from the maximum value in the list. The input begins with an integer indicating the number of integers that follow.

in python please

Write a version of the binary search algorithm that can be used to search a list of strings. (Use the selection sort that you designed to sort the list.)

PYTHON Complete the function below, which takes two arguments: data: a list of tweets search_words: a list of search phrases The function should, for each tweet in data, check whether that tweet uses any of the words in the list search_words. If it does, we keep the tweet. If it does not, we ignore the tweet. data = ['ZOOM earnings for Q1 are up 5%', 'Subscriptions at ZOOM have risen to all-time highs, boosting sales', "Got a new Mazda, ZOOM ZOOM Y'ALL!", 'I hate getting up at 8am FOR A STUPID ZOOM MEETING', 'ZOOM execs hint at a decline in earnings following a capital expansion program'] Hint: Consider the example_function below. It takes a list of numbers in numbers and keeps only those that appear in search_numbers. def example_function(numbers, search_numbers): keep = [] for number in numbers: if number in search_numbers(): keep.append(number) return keep def search_words(data, search_words):

The function count_contains_og in python takes a list of strings and returns how many strings in the list contain 'og' / 'OG' / 'oG' / 'Og' (check for 'og', ignoring case). Hint: Use the sequence membership operator in to help you check for 'og' in the individual strings. Create a lower-cased version of the string (lower), then use the in operator. For example: Test Result str_list = ['cat', 'dog', 'FROG', 'monkey'] print(count_contains_og(str_list)) 2 strlist = ["X", "x"] print(count_contains_og(strlist)) 0 list_of_one_og = ["Doggie"] print(count_contains_og(list_of_one_og)) 1

IN PYTHON THANK YOU

Assume, you have been given two lists: List_one and List_two. [Your program should work for any two given lists; change the following lists and check whether your program works correctly for the code you have written] Write a Python program that prints "True", if the given two lists have at least one common member. Otherwise, print "False". Hint: use a boolean variable as a flag to indicate if the two lists have at least one common element. Use break to end the loop when seeing a commom element. =================================================================== Given lists 1:List_one : [1, 4, 3, 2, 6]List_two : [5, 6, 9, 8, 7] Sample Output 1:True =================================================================== Given lists 2:List_one : [1, 4, 3, 2, 5]List_two : [8, 7, 6, 9] Sample Output 2:False #assign the boolean result (True/False) to variable "common_ele" (flag). def task5(list_1, list_2): # YOUR CODE HERE return common_ele Expert Solution

A sequential search of a sorted list can halt when the target is less than a given element in the list. Modify the program to stop when the target becomes less than the current value being compared. In a sorted list, this would indicate that the target is not in the list and searching the remaining values is unnecessary. ----------------------------------------------------------------------------------- """ File: search.py Project 11.1 """ def sequentialSearch(target, lyst): """Returns the position of the target item if found, or -1 otherwise. The lyst is assumed to be sorted in ascending order.""" position = 0 while position < len(lyst): if target == lyst[position]: return position position += 1 return -1 def main(): """Tests with three lists.""" print(sequentialSearch(3, [0, 1, 2, 3, 4])) print(sequentialSearch(3, [0, 1, 2])) # Should stop at second position. print(sequentialSearch(3, [0, 4, 5, 6])) if...