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1f) Percent Difficult Words Using dale_chall_words, define a function percent_difficult_words that given a text input will calculate the percent difficult words, using the following approach: Return the percent of words in a given text input that are not in dale_chall_words['easy_words']. Note that a "word" is defined by being separated from other text by a space.

1f) Percent Difficult Words Using dale_chall_words, define a function percent_difficult_words that given a text input will calculate the percent difficult words, using the following approach: Return the percent of words in a given text input that are not in dale_chall_words['easy_words']. Note that a "word" is defined by being separated from other text by a space.

Database System Concepts
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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1f) Percent Difficult Words Using dale_chall_words, define a function percent_difficult_words that given a text input will calculate the percent difficult words, using the following approach: Return the percent of words in a given text input that are not in dale_chall_words['easy_words']. Note that a "word" is defined by being separated from other text by a space. I # YOUR CODE HERE def percent_difficult_words(dale_chall_file): text_input text_input_words = text_input.split() difficult_words_count = 0 input ("Please Enter a text : ") for i in text_input_words: if i not in dale_chall_file: difficult_words_count += 1 return difficult_words_count / len(text_input_words) * 100 final_percent_difficult_words print("Difficult words percentage is : = percent_difficult_words(dale_chall_file) ,final_percent_difficult_words, "%") Please Enter a text : there are only easy words here Difficult words percentage is : 100.0 % 1g) Calculate Percent Difficult words Apply the above method to 'message' column of news_df , creating a new column in news_df called 'PDW' . This column should store the percent difficult words of each message. I # YOUR CODE HERE I assert news_df['PDW']. max() assert news_df['PDW'].min() == 1.0 == 0 assert round(news_df['PDW'].mean(), 2) == 0.48 I # Look at output news_df.head()
Transcribed Image Text:1f) Percent Difficult Words Using dale_chall_words, define a function percent_difficult_words that given a text input will calculate the percent difficult words, using the following approach: Return the percent of words in a given text input that are not in dale_chall_words['easy_words']. Note that a "word" is defined by being separated from other text by a space. I # YOUR CODE HERE def percent_difficult_words(dale_chall_file): text_input text_input_words = text_input.split() difficult_words_count = 0 input ("Please Enter a text : ") for i in text_input_words: if i not in dale_chall_file: difficult_words_count += 1 return difficult_words_count / len(text_input_words) * 100 final_percent_difficult_words print("Difficult words percentage is : = percent_difficult_words(dale_chall_file) ,final_percent_difficult_words, "%") Please Enter a text : there are only easy words here Difficult words percentage is : 100.0 % 1g) Calculate Percent Difficult words Apply the above method to 'message' column of news_df , creating a new column in news_df called 'PDW' . This column should store the percent difficult words of each message. I # YOUR CODE HERE I assert news_df['PDW']. max() assert news_df['PDW'].min() == 1.0 == 0 assert round(news_df['PDW'].mean(), 2) == 0.48 I # Look at output news_df.head()
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