mecha... 31 Running Tips To... 9 029-Allens, Ashur.... 5 Forsyth Street, O'... Talia Simmons Write a function inverted_w

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Mecha 31 Running Tips To 9 029 Allens Ashur 5 Forsyth Street O Talia Simmons Write A Function Inverted W 1 (181.44 KiB) Viewed 65 times

mecha... 31 Running Tips To... 9 029-Allens, Ashur.... 5 Forsyth Street, O'... Talia Simmons Write a function inverted_word_counts(word_count_dict, banned) that takes as a parameter a 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])) L Answer: (penalty regime: 0, 10, 20, ... %) X 644= 330KG₂ 744 370KES Boks 240ke $347 Prax Search Results | Bur... 19°C Mostly cloudy 3:1 ENG 9/06