Suppose 'start' : 1472 is an element in a dictionary. What is the key? What isthe value?
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Suppose 'start' : 1472 is an element in a dictionary. What is the key? What is
the value?
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- To create a dictionary, you just surround a sequence of value pairs of separated by: and the pair separated by commas, within curly braces. True or false? How can you determine whether a key-value pair exists in a dictionary?The function below, get_value_if_key, takes two arguments: the dictionary data_dict and a key of any type my_key. Fix the code to return the value from data_dict for key my_key if that key exists. The function should not return anything otherwise (i.e. return None).
- Create a dictionary (order_dict) that the key is the burger number, and the value is a list of the quantities. For example: order_dict which is {'1': [10, 3, 78], '2': [35, 0, 65], '3': [9, 23, 0], '4': [0, 19, 43], '5': [43, 0, 21]} Create a dictionary (total_order_dict) that the key is the burger number, and the value is the total of the quantities for that burger. For example: total_order_dict which is {'1': 91, '2': 100, '3': 32, '4': 62, '5': 64} Use "assert to test the values of the total_order_dict For example: expected_result_ototal_order_dict = {'1': 91, '2': 100, '3': 32, '4': 62, '5': 64} assert actual_result_ototal_order_dict['1'] == expected_result_ototal_order_dict['1'] , "The actual result is not the \same as expected result for the order number 1!" (and test the other elements) The function can be like: def test_sum(): #Testing assert actual_result_ototal_order_dict['1'] == 91, "The actual result is not the \ same as expected result for the…How do you create an empty dictionary?def 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',…
- Create a static dictionary with a number of users and with the following values: First name Last name Email address Password Ask the user for: 5. Email address 6. Password Loop (for()) through the dictionary and if (if()) the user is found print the following: 7. Hello, first name last name you have successfully logged in 8. Notify the user if the password and email address are wrong 9. Additional challenge: if you want the program to keep asking for a username and password when the combination is wrong, you will need a while() loop. 10. Save the file as assignment03yourlastname.pydef add_key_values(x: str)-> dict[str, int]: "Return a dictionary where the keys are the distinct characters appearing in x and the values are how many times they appear." Starting from the top and considering each testcase in turn, please mark the 6 options that add nothing to what was covered by the previous tests. x = "", return: {} x = "a", return {"a":1} x="ab", return {["a": 1, "b":1} x = "aab", return {"a": 2, "b": 1} x = "1233", return: {"1": 1, "2": 1, "3":2} x = "1223334444", return: {"1": 1, "2": 2, "3": 3, "4":4} x = "cabba", return: {"c": 1, "a": 2, "b":2} x = "qnwhnwq", return: {"q": 2, "n": 2, "w": 2, "h": 1} x = "a,aa33b", return {"a": 3, ",": 1, "3": 2, "b":1} x = "midterm", return {"m": 2, "i": 1, "d": 1, "t": 1, "e": 1, "r": 1} x = "csc108ishard", return {"c": 2, "s": 2, "1": 1, "0": 1, "8: 1, "i": 1, "h": 1, "a": 1, "": 1, "d":1}how to display the only the value of a dictionary in upper case
- Problem The program takes a dictionary and checks if a giver. key exists in a dictionary or not.Question 13: Create an empty dictionary as an object of dictionary class. Add four key-value pairs to that empty dictionary one-by-one. Then, remove one of the pairs and display the latest version of the dictionary. use type () function.Please answer the ques in python with showing the code