inverted_word_counts(word_count_dict, banned) that takes as a parameter a Write a function dictionary mapping from words to word-counts (like the ones in earlier questions) and a list of banned words, then returns the dictionary inverted, i.e. mapping from an integer word count to an alphabetically ordered list of words with that count but skipping words that are banned. For example, if the word count dictionary were: {'a': 20, 'west': 10, 'blotto': 1, 'bingo': 1, 'x': 5, 'member': 1) and the banned list = ["member", "west"] then the inverted dictionary would be (1: ['bingo', 'blotto'], 5: ['x'], 20: ['a']} showing that the list of words occurring exactly once is ['bingo', 'blotto] (note that it's alphabetically ordered), the list of words occurring five times is just 'x' and the list of words occurring 20 times is ['a']. For example: Result Test test_dict= {'a': 20, 'west': 10, 'blotto': 1, "bingo': 1, "x": 5, "member: 1) 1: ['bing banned - ["member", "west"] 5: [***] 20: ['a'] inverse = inverted word_counts (test_dict, banned) for count in sorted (inverse.keys()): print('(): ().format(count, inverse[count])) Answer: (penalty regime: 0, 10, 20, ... %) 1

inverted_word_counts(word_count_dict, banned) that takes as a parameter a Write a function dictionary mapping from words to word-counts (like the ones in earlier questions) and a list of banned words, then returns the dictionary inverted, i.e. mapping from an integer word count to an alphabetically ordered list of words with that count but skipping words that are banned. For example, if the word count dictionary were: {'a': 20, 'west': 10, 'blotto': 1, 'bingo': 1, 'x': 5, 'member': 1) and the banned list = ["member", "west"] then the inverted dictionary would be (1: ['bingo', 'blotto'], 5: ['x'], 20: ['a']} showing that the list of words occurring exactly once is ['bingo', 'blotto] (note that it's alphabetically ordered), the list of words occurring five times is just 'x' and the list of words occurring 20 times is ['a']. For example: Result Test test_dict= {'a': 20, 'west': 10, 'blotto': 1, "bingo': 1, "x": 5, "member: 1) 1: ['bing banned - ["member", "west"] 5: [***] 20: ['a'] inverse = inverted word_counts (test_dict, banned) for count in sorted (inverse.keys()): print('(): ().format(count, inverse[count])) Answer: (penalty regime: 0, 10, 20, ... %) 1

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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