lines = x.readlines() for line in lines: if line.find(word) > -1: n+=1 return n SRP (single responsibility principle) 1. Break up main problem into smaller logical sub problems. 2. solve each sub problem. 3. Use glue code helps join sub problems into a functional solution. main() 1. read file and return all lines 2. tokenize each line and create histogram for each word in the line. h = { 'this': [1, [filename]], 'is':[1,filename], 'my':[1, filename], 'car': [1, filename] } h = {'this':1, 'is':1, 'my':1, 'truck': 1} a = {'this':[2, [filename1, filename2]], 'is':2, [filename1, filename2]], 'my':2, 'car': 1, 'truck':1} 3. aggregate histograms 4. glue code: look for cars and trucks k = ['cars', 'trucks'] cars = [] tucks = 0] if k[0] in a.keys(): cars += (a[i][1]) if k[1] in a.keys(): trucks += (a[i][1])
lines = x.readlines() for line in lines: if line.find(word) > -1: n+=1 return n SRP (single responsibility principle) 1. Break up main problem into smaller logical sub problems. 2. solve each sub problem. 3. Use glue code helps join sub problems into a functional solution. main() 1. read file and return all lines 2. tokenize each line and create histogram for each word in the line. h = { 'this': [1, [filename]], 'is':[1,filename], 'my':[1, filename], 'car': [1, filename] } h = {'this':1, 'is':1, 'my':1, 'truck': 1} a = {'this':[2, [filename1, filename2]], 'is':2, [filename1, filename2]], 'my':2, 'car': 1, 'truck':1} 3. aggregate histograms 4. glue code: look for cars and trucks k = ['cars', 'trucks'] cars = [] tucks = 0] if k[0] in a.keys(): cars += (a[i][1]) if k[1] in a.keys(): trucks += (a[i][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
Related questions
Question
I need help with this python
![lines = x.readlines()
for line in lines:
if line.find(word) > -1:
n+=1
return n
SRP (single responsibility principle)
1. Break up main problem into smaller logical sub problems.
2. solve each sub problem.
3. Use glue code helps join sub problems into a functional solution.
main()
1. read file and return all lines
2. tokenize each line and create histogram for each word in the line.
h = { 'this':[1, [filename]], 'is':[1,filename], 'my':[1, filename], 'car': [1, filename] }
h = {'this':1, 'is': 1, 'my':1, 'truck': 1}
a = {'this':[2, [filename1, filename2]], 'is':2, [filename1, filename2]], 'my':2, 'car':1, 'truck':1}
3. aggregate histograms
4. glue code:
look for cars and trucks
k = ['cars', 'trucks']
cars = []
tucks = []
if k[0] in a.keys():
cars += (a[i][1])
if k[1] in a.keys():
trucks += (a[i][1])](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F00545010-2a64-4380-b210-1d7051010133%2Fdd3a7347-41f1-45d8-ae19-4fb2d0ffa2f5%2F6gg4jhh_processed.png&w=3840&q=75)
Transcribed Image Text:lines = x.readlines()
for line in lines:
if line.find(word) > -1:
n+=1
return n
SRP (single responsibility principle)
1. Break up main problem into smaller logical sub problems.
2. solve each sub problem.
3. Use glue code helps join sub problems into a functional solution.
main()
1. read file and return all lines
2. tokenize each line and create histogram for each word in the line.
h = { 'this':[1, [filename]], 'is':[1,filename], 'my':[1, filename], 'car': [1, filename] }
h = {'this':1, 'is': 1, 'my':1, 'truck': 1}
a = {'this':[2, [filename1, filename2]], 'is':2, [filename1, filename2]], 'my':2, 'car':1, 'truck':1}
3. aggregate histograms
4. glue code:
look for cars and trucks
k = ['cars', 'trucks']
cars = []
tucks = []
if k[0] in a.keys():
cars += (a[i][1])
if k[1] in a.keys():
trucks += (a[i][1])
![De Morgan's laws commonly apply to text searching using Boolean operators AND, OR, and NOT. Consider a set of
documents containing the words "cars" and "trucks". De Morgan's laws hold that these two searches will return the
same set of documents:
Search A: NOT (cars OR trucks)
Search B: (NOT cars) AND (NOT trucks)
The corpus of documents containing "cars" or "trucks" can be represented by four documents:
Document 1: Contains only the word "cars".
Document 2: Contains only "trucks".
Document 3: Contains both "cars" and "trucks".
Document 4: Contains neither "cars" nor "trucks".
To evaluate Search A, clearly the search "(cars OR trucks)" will hit on Documents 1, 2, and 3. So the negation of that
search (which is Search A) will hit everything else, which is Document 4.
Evaluating Search B, the search "(NOT cars)" will hit on documents that do not contain "cars", which is Documents 2
and 4. Similarly the search "(NOT trucks)" will hit on Documents 1 and 4. Applying the AND operator to these two
searches (which is Search B) will hit on the documents that are common to these two searches, which is Document 4.
Please write a Python program to De Morgan's Law via document search.
You can use four text files as document 1, 2 3 and 4.
Search A: NOT (cars OR trucks)
set A= list of all docs (a, b, c, d)
set B = list of docs with trucks (a, c)
set C = list of docs with cars (b, c)
set D = list of docs with cars or trucks (a, b, c)
set E = not D (d)
Search B: (NOT cars) AND (NOT trucks)
set A
list of all docs (a, b, c, d)
set B
not cars (a,d)
set C = not trucks (b,d)
set D = (not Cars AND not Trucks) (d)
d = {}
d['cars'] = [a, c]
d['trucks'] = [b, c]
def get_word_count(word, file):
n = 0
x = open(..)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F00545010-2a64-4380-b210-1d7051010133%2Fdd3a7347-41f1-45d8-ae19-4fb2d0ffa2f5%2Fpp4bi7o_processed.png&w=3840&q=75)
Transcribed Image Text:De Morgan's laws commonly apply to text searching using Boolean operators AND, OR, and NOT. Consider a set of
documents containing the words "cars" and "trucks". De Morgan's laws hold that these two searches will return the
same set of documents:
Search A: NOT (cars OR trucks)
Search B: (NOT cars) AND (NOT trucks)
The corpus of documents containing "cars" or "trucks" can be represented by four documents:
Document 1: Contains only the word "cars".
Document 2: Contains only "trucks".
Document 3: Contains both "cars" and "trucks".
Document 4: Contains neither "cars" nor "trucks".
To evaluate Search A, clearly the search "(cars OR trucks)" will hit on Documents 1, 2, and 3. So the negation of that
search (which is Search A) will hit everything else, which is Document 4.
Evaluating Search B, the search "(NOT cars)" will hit on documents that do not contain "cars", which is Documents 2
and 4. Similarly the search "(NOT trucks)" will hit on Documents 1 and 4. Applying the AND operator to these two
searches (which is Search B) will hit on the documents that are common to these two searches, which is Document 4.
Please write a Python program to De Morgan's Law via document search.
You can use four text files as document 1, 2 3 and 4.
Search A: NOT (cars OR trucks)
set A= list of all docs (a, b, c, d)
set B = list of docs with trucks (a, c)
set C = list of docs with cars (b, c)
set D = list of docs with cars or trucks (a, b, c)
set E = not D (d)
Search B: (NOT cars) AND (NOT trucks)
set A
list of all docs (a, b, c, d)
set B
not cars (a,d)
set C = not trucks (b,d)
set D = (not Cars AND not Trucks) (d)
d = {}
d['cars'] = [a, c]
d['trucks'] = [b, c]
def get_word_count(word, file):
n = 0
x = open(..)
Expert Solution
Step 1
Python:-
In python procedural- and object-oriented programming are supported. Because Python is a dynamically typed language, we don't need to declare the type of a variable. As an illustration, x = 10. In this case, x can be anything, including a String, an int, etc.
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