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
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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Question
Using Python, create a Percent Difficult Words column using the method above
![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()](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3200e892-7b83-4670-8aa5-c4c84f2a6adb%2F317261ca-0a7a-4f9d-a4a1-d1d9e315f0bd%2Fa664cp_processed.png&w=3840&q=75)
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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