swift WAP for a dictionary is given print all values with key , then add one more entry .now print new dictionary and then change a value using key then again print new dictionary
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WAP for a dictionary is given print all values with key , then add one more entry .now print new dictionary and then change a value using key then again print new dictionary
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- For example, suppose the word “robot” is found in lines 7, 18, 94, and 138. The dictionary would contain an element in which the key was the string “robot”, and the value was a list containing the numbers 7, 18, 94, and 138. Once the dictionary is built, you should prompt the user for a word. If the word is found in the dictionary, then print out all the lines on which they are found in ascending order. The code for the print out is provided in the stub code. A stub code is provided for you. If you do not see it, click the "Load default template" button. # The get_word_dict function returns a dictionary containing# the words from line_list as keys, and their line numbers# as values.def get_word_dict(line_list):# Create a line counter.count = 0 # Create a dictionary to hold the words.word_dict = {} # Step through the list of lines.#######Add you code below############# ####################################### Return the dictionary.return word_dict #####DO NOT CHANGE ANYTHING…Counting numbers are the numbers 1, 2, 3, and so on. Create a dictionary in Python that maps the first n counting numbers to their squares. Assign the dictionary to the variable squares.From two dictionary objects created in the codes below, find letters that only exist in one of the dictionaries. Use a SET object(s) and its mathematical operator(s). import string # import string library import random # import random library # a_base is a string containing concatenation of letters s_base = string.ascii_letters # random.seed() accepts any integer to initialize random numbers random.seed(582) # s is a string variable s = "" # this line creates a loop starting from 0 to 199 for i in range(200): # this line randomly selects letters from s_base variable and concatenates them to the value stored in s s += s_base[random.randint(0, 25)] # this line prints the value of s print(s) # count each char frequency count = {} for ch in s: if ch in count: count[ch] = count[ch]+1 else: count[ch] = 1 # print the frequency for key in count: print("{} -> {}".format(key, count[key])) def char_frequency(s): dict = {} for d in s: keys =…
- In python: student_dict is a dictionary with students' name and pedometer reading pairs. A new student is read from input and added into student_dict. For each student in student_dict, output the student's name, followed by "'s pedometer reading: ", and the student's pedometer reading. Then, assign average_value with the average of all the pedometer readings in student_dict..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 puthondef city_dict(adict): """ Question 6 -Given a dictionary that maps a person to a list of countries they want to travel to, return a dictionary with the value being the list sorted by the last letter in each countries name. If two countries have the same last letter, sort by the first letter. -For an extra challenge, try doing this in one line (Optional). Args: adict (dict) Returns: dict >>> city_dict({"Pablo": ["Belgium", "Canada", "Germany"], "Athena": ["Italy", "France", "Egypt"], "Liv": ["Japan", "Bolivia", "Greece"]}) {'Pablo': ['Canada', 'Belgium', 'Germany'], 'Athena': ['France', 'Egypt', 'Italy'], 'Liv': ['Bolivia', 'Greece', 'Japan']} >>> city_dict({"Jacob": ["Bahamas", "Brazil", "Chile"], "Lexi": ["Colombia", "Finland", "Panama"], "Emily": ["Ireland", "Russia", "Kenya", "Jordan"]}) {'Jacob': ['Chile', 'Brazil', 'Bahamas'], 'Lexi': ['Colombia', 'Panama', 'Finland'], 'Emily': ['Kenya',…
- Now do it the fixed sliding window of building a tri-gram dictionary method. You can start after 256 for the dictionary index. (a) Use window size 3 for the text: “pugonthepugonrug” (b) Think about how many of these tri-grams are used (or not used). Can you think of a way to reduce the number of tri-grams that only occur once?Assignment 1: Prompt the user for 5 pairs of numbers. The pairs consist of a player’s jersey number (0-99) and the player’s rating (1-9). Make sure to use good prompts and check the user’s input using a while loop. Do not stop the program if the input is outside the range. You need to prompt the user until they get the numbers correct. Store the pairs in a dictionary. Since the player’s jersey number is a key, you need to check for duplicates and prompt user again if the number is already in use. After all players are entered, print the roster with the jersey numbers in ascending order. Next, print a menu for the user to be able to modify the roster. They should be able to add a new player, remove a player, update a player’s rating, output a list of players above a rating (get the cutoff from the user), output the roster, or quit. You need to turn in a written algorithm for this project before starting to write the program. Assignment 2: Change assignment 1 to use functions.…You will be given a dictionary, where the name of a student will be the key and a list of courses will be the value. Now, you have to take a course name as input until the user gives 'STOP' in the input. Your task is to display the list of the students taken that course after each input. Given Dictionary { 'Mike' : ['CSE110","ENG101','MAT110'], "Simon' : ['CSE111',"PHY111',"MAT110", 'CSE230], "Shawn' : ['CSE110',"ENG101',"MAT120','CSE230'], 'Alice' : ['CSE110',"ENG091','MAT092'] } Sample Input 1 CSE110 МAT110 CSE260 STOP Sample Output 1 [ 'Mike', "Shawn', 'Alice' ] [ 'Mike', "Simon'] [] Explanation 1: Mike, Shawn and Alice are doing CSE110 Mike and Simon are doing MAT110 No student is doing CSE260
- Creates and returns a mutable dictionary, initially giving it enough allocated memory to hold a given number of entries.Creates and returns a mutable dictionary, initially giving it enough allocated memory to hold a given number ofentries.That's enough of you! def remove_after_kth(items, k=1): Given a list of items, some of which may be duplicated, create and return a new list that is otherwise the same as items, but only up to k occurrences of each element are kept, and all occurrences of that element after those first k are discarded.Hint: loop through the items, maintaining a dictionary that remembers how many times you have already seen each element. Update this count as you go, and append each element to the result list only if its count is still at most equal to k. items k Expected result [42, 42, 42, 42, 42, 42, 42] 3 [42, 42, 42] ['tom', 42, 'bob', 'bob', 99, 'bob', 'tom', 'tom', 99] 2 ['tom', 42, 'bob', 'bob', 99, 'tom', 99] [1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1] 1 [1, 2, 3, 4, 5] [1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5] 3 [1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 1, 5] [42, 42, 42, 99, 99, 17] 0 []