screenshot of input and output import stringpp = '''Peter Piper picked a peck of pickled peppers.A peck of pickled peppers Peter Piper picked.If Peter Piper picked a peck of pickled peppersWhere's the peck of pickled peppers Peter Piper picked?'''stopwords = "a,an,any,can,if,is,it,it's,nor,not,of,on,nor,the"def remove_punc(s):'''Returns a copy of string s with all punctuation removed'''#This function is already written - don't change it!for char in string.punctuation:s = s.replace(char, '')return sdef count_words(text, stopwords):'''Returns a dictionary with words and their counts'''pass#Create an empty dictionary#Split the text into a list of words (split on whitespace)#Split the stopwords into another list of words (split on comma)#For each word in the list#If the word is not in the stop list#If the word is not in the dictionary#Add it with a count of 1#Otherwise#Add 1 to the count#Return the dictionarydef show_counts(word_dict):'''Display words and their counts'''pass#Get a list of the keys from word_dict#Put the words in alphabetical order#For each word in the list#Print the word and its countdef main():'''Controls the program'''text = remove_punc(pp.lower()) #text is the name of the string with no punctuation#Use count_words to get a dictionary with words and counts; send it text and stopwords#Use show_counts() to display words and countsmain() #If the word is not in the stop list#If the word is not in the dictionary242526#Add it with a count of 127#Otherwise28#Add 1 to the count29#Return the dictionary3031 def show_counts (word_dict) :32'''Display words and their counts'''33pass34#Get a list of the keys from word_dict35#Put the words in alphabetical order36#For each word in the list37#Print the word and its count3839 def main () :40'''Controls the program'''41text = remove_punc (pp.lower ())ttext is the name of the string42#Use count_words to get a dictionary with words and counts; sen43#Use show_counts () to display words and counts4445 main ()46 3. Often, when dealing with text, we want to ignore certain words. For example, when searching fortext we probably don't want to include words like a, the, of, and so forth. These words are called'stopwords' and are usually kept in a file. Write a program to count the number of occurrences ofwords in some text but skip the stopwords. Again, use a dictionary to count the occurrences thendisplay the words (in alphabetical order) and their counts. Use the text in the variable pp - but yourprogram should work for any text; the stopwords are stored in the variable stopwords. Note that thestopwords are separated by comma - not space.Save & RunLoad History1 import string23 pp = '''Peter Piper picked a peck of pickled peppers.4 A peck of pickled peppers Peter Piper picked.5 If Peter Piper picked a peck of pickled peppers6 Where's the peck of pickled peppers Peter Piper picked?''"78 stopwords = "a, an, any, can, if, is, it,it's,nor, not, of, on, nor, the"10 def remove_punc (s):11'''Returns a copy of string s with all punctuation removed'''12#This function is already written - don't change it!13for char in string.punctuation:14= = s.replace (char, '')15return S1617 def count_words (text, stopwords) :18'''Returns a dictionary with words and their counts'''19pass20#Create an empty dictionary#Split the text into a list of words (split on whitespace)#Split the stopwords into another list of words (split on comma#For each word in the list212223

keys() is a method in dictionaries which will return us a list of dictionary keys sorted() sorts…

screenshot of input and output import string pp = '''Peter Piper picked a peck of pickled peppers. A peck of pickled peppers Peter Piper picked. If Peter Piper picked a peck of pickled peppers Where's the peck of pickled peppers Peter Piper picked?''' stopwords = "a,an,any,can,if,is,it,it's,nor,not,of,on,nor,the" def remove_punc(s): '''Returns a copy of string s with all punctuation removed''' #This function is already written - don't change it! for char in string.punctuation: s = s.replace(char, '') return s def count_words(text, stopwords): '''Returns a dictionary with words and their counts''' pass #Create an empty dictionary #Split the text into a list of words (split on whitespace) #Split the stopwords into another list of words (split on comma) #For each word in the list #If the word is not in the stop list #If the word is not in the dictionary #Add it with a count of 1 #Otherwise #Add 1 to the count #Return the dictionary def show_counts(word_dict): '''Display words and their counts''' pass #Get a list of the keys from word_dict #Put the words in alphabetical order #For each word in the list #Print the word and its count def main(): '''Controls the program''' text = remove_punc(pp.lower()) #text is the name of the string with no punctuation #Use count_words to get a dictionary with words and counts; send it text and stopwords #Use show_counts() to display words and counts main()

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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Question

screenshot of input and output

import string

pp = '''Peter Piper picked a peck of pickled peppers.
A peck of pickled peppers Peter Piper picked.
If Peter Piper picked a peck of pickled peppers
Where's the peck of pickled peppers Peter Piper picked?'''

stopwords = "a,an,any,can,if,is,it,it's,nor,not,of,on,nor,the"

def remove_punc(s):
'''Returns a copy of string s with all punctuation removed'''
#This function is already written - don't change it!
for char in string.punctuation:
s = s.replace(char, '')
return s

def count_words(text, stopwords):
'''Returns a dictionary with words and their counts'''
pass
#Create an empty dictionary
#Split the text into a list of words (split on whitespace)
#Split the stopwords into another list of words (split on comma)
#For each word in the list
#If the word is not in the stop list
#If the word is not in the dictionary
#Add it with a count of 1
#Otherwise
#Add 1 to the count
#Return the dictionary

def show_counts(word_dict):
'''Display words and their counts'''
pass
#Get a list of the keys from word_dict
#Put the words in alphabetical order
#For each word in the list
#Print the word and its count

def main():
'''Controls the program'''
text = remove_punc(pp.lower()) #text is the name of the string with no punctuation
#Use count_words to get a dictionary with words and counts; send it text and stopwords
#Use show_counts() to display words and counts

main()