def week_10(report): k= [] for i in report: Z = for j in range(i.index("bio"), len(i)): if ilj].isdigit(): z += ilil elif ilj] == ": z += ilil k.append(z) return k
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Can this be put into a list comprehension? (On one line)??
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- An indexer, which functions similarly to an array indexer, may be used to get a value using its associated key. As the index value, a key is provided in, and the value associated with the key is returned, unless the key does not exist, in which case a null is returned.use c#This approach is demonstrated by the following short code segment:Implement a city database using ordered lists by using java. Each database record contains the name of the city (a string of arbitrary length) and the coordinates of the city expressed as integer x and y coordinates. Your database should allow records to be inserted, deleted by name, and searched by name. Another operation that should be supported is to print all records within a given distance of a specified point/coordinate. The order of cities should be alphabetically by city name. Implement the database using both: an arraybased list implementation, and a circular single linked list implementation.Use may the following node, SLL implementations to implement an ordered circular single linked listdef 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 """
- Implement a city database using ordered lists. Each database record contains the name of the city (a string of arbitrary length) and the coordinates of the city expressed as integer x and y coordinates. Your database should allow records to be inserted, deleted by name, and searched by name. Another operation that should be supported is to print all records within a given distance of a specified point/coordinate. The order of cities should be alphabetically by city name. Implement the database using: an array- based list implementation. By using JAVA.Suppose a list contains the final exam results of CSE110, PHY111, and MAT110 of each student in a listed form. Your task is to sort the list based on a course name that will be given by the user and print the names after sorting. For example, a list looks likelst = [ ["Farzad",95,87,91] , ["Mahjabin",92,90,83], ["Sadman",87,92,80], ["Nakhla",85,94,90] ]where for each list, 1st index is name, 2nd index is CSE110, 3rd index is PHY111 and 4th index is MAT110. ===================================================== Hint:create a python function for sorting, then call it everytime instead of rewriting the code. Get the data in the individual lists from the given nested list. ===================================================== Sample InputMAT110 Sample OutputFarzadNakhlaMahjabinSadmanI was wondering if you could describe the difference between a clustered index and a sparse index.
- Python Machine Learning Can you tell me what are the outliners in this dataset? from sklearn.datasets import load_wine print(load_wine().DESCR) X = load_wine().data y = load_wine().target f_names = load_wine().feature_names import pandas as pd df = pd.DataFrame(X, columns=f_names) df['TARGET'] = y df.head() df['Type'] = y df.head() df.info() df.isna().sum().sum() df.describe() import matplotlib.pyplot as plt df.hist(figsize=(20,12), layout=(4,8)) plt.tight_layoutAnswer the given question with a proper explanation and step-by-step solution. Java N people are arranged in a circle.(Each assigned a number in sequence). Starting from the first, countdown , removing each k-th person from the list, while closing the circle. Determine the number of the last one remaining in the circle (k<=N) Hint: to solve this problem, use a list in which the reference field of the last element contains a link to the first elementIf you had the option of having a million-element list, which implementation of the size and isEmpty functions would you use? (Consider the issue with the University of Michigan's alumni database.) If you had the option of making a million little lists, how would you choose? Think about the challenge of tracking dependents for a million income tax filings.)
- Which implementation of the size and isEmpty methods would you use if you had the potential for a million-element list. (Consider the problem of keeping track of the alumni for the University of Michigan.) How would you choose if you had the potential for a million small lists. (Consider the problem of keeping track of the dependents for each of a million income-tax returns.)Computer ScienceA restaurant requires WAIT ADT to help waiter to wait tables. Theapplication is required to provide functions to add a food item to the WAITlist, remove an item from the Wait list, size of the list i.e., the food itemsordered, and know if item is already in the list. The WAIT ADT access timeshould be O [1]. Provide implementation details with your choice of datastructure and the operations with justification of the constant access time.Implement search food in the added Wait list in C.Data Algorithms (language: Java) Describe what each function does and its Big O? import java.util.ArrayList;import java.util.Arrays;import java.util.List;import java.util.*;public class HW04 {public static boolean checkPrimeNumber(int n){for(int i = 2; i <= n / 2; i++)if(n % i == 0)return false;return true;}public static List<List<Integer>> multiSum(int n) {if(n <= 0){throw new NumberFormatException("Not supported yet.");}List<List<Integer>> p = new ArrayList<>();p.add(Arrays.asList(1,n));for(int i = n - 1; i >= Math.sqrt(n); i--){if(n % i == 0){int left = n / i;p.add(Arrays.asList(i, left));if(! checkPrimeNumber(i)) {List<List<Integer>> lst = multiSum(i);for(int k = 1; k < lst.size();k++) {List<Integer> currList = new ArrayList<>();currList.add(left);for(int j = 0; j < lst.get(k).size(); j++) {currList.add(lst.get(k).get(j));}p.add(currList);}}}}for(int i = 0; i < p.size(); i++)…