2. Repeat the previous problem but instead of a query word, input a query number (a threshold) and return a Python dictionary that contains the words with its corresponding counts exceeding or equal to the query number In [ ]: import string # import string for punctuation characters def commonwords (filename, threshold): if type (filename) is not str: raise TypeError("First arguments must be a string") if type (threshold) is not int: raise TypeError("second arguments must be an integer") # YOUR CODE HERE raise NotImplementedError() In [ ]: # DO NOT MODIFY, THIS IS UNIT TESTING! [1 pt] from nose.tools import assert_equal, assert_is_instance assert_equal (commonwords ("alice29.txt", see), ('the': 1514, 'to': 717, 'and': 774, 'a': 610)) assert equal (commonwords ("alice29.txt", 35e), ('I': 401, 'the': 1514, 'Alice': 385, was': 352, 'to': 717, 'of': 493, 'and': 774, 'she': 497, 'it': 482, 'in': 35 assert equal(commonwords ("alice29.txt", 200), ('I': 401, 'the': 1514, 'Alice': 385, was: 352, 'to': 717, 'of': 493, 'her': 243, 'and': 774, 'she': 497, "it': 4 assert_is_instance (commonilords ("alice29.txt", 1000), dict)

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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2. Repeat the previous problem but instead of a query word, input a query number (a threshold) and return a Python dictionary that contains the words with
its corresponding counts exceeding or equal to the query number
In [ ]: import string # import string for punctuation characters
def commonwords (filename, threshold):
if type(filename) is not str:
raise TypeError("First arguments must be a string")
if type (threshold) is not int:
raise TypeError("Second arguments must be an integer")
# YOUR CODE HERE
raise NotImplementedError()
In [ ]: # DO NOT MODIFY, THIS IS UNIT TESTING! [1 pt]
from nose.tools import assert_equal, assert_is_instance
assert equal (commonwords ("alice29.txt", see), ('the': 1514, "to': 717, 'and': 774, 'a': 610})
assert equal (commonwords ("alice29.txt", 350),
('I': 401, 'the': 1514, 'Alice': 385, 'was': 352, 'to': 717, 'of': 493, "and': 774, 'she': 497, 'it': 482, 'in': 35)
assert equal (commonwords ("alice29.txt", 200),
('I': 401, 'the': 1514, 'Alice': 385, 'Nas': 352, 'to': 717, 'of': 493, 'her': 243, 'and': 774, "she': 497, "it': 4
assert_is_instance (connonwords("alice29.txt", 1000), dict)
Transcribed Image Text:2. Repeat the previous problem but instead of a query word, input a query number (a threshold) and return a Python dictionary that contains the words with its corresponding counts exceeding or equal to the query number In [ ]: import string # import string for punctuation characters def commonwords (filename, threshold): if type(filename) is not str: raise TypeError("First arguments must be a string") if type (threshold) is not int: raise TypeError("Second arguments must be an integer") # YOUR CODE HERE raise NotImplementedError() In [ ]: # DO NOT MODIFY, THIS IS UNIT TESTING! [1 pt] from nose.tools import assert_equal, assert_is_instance assert equal (commonwords ("alice29.txt", see), ('the': 1514, "to': 717, 'and': 774, 'a': 610}) assert equal (commonwords ("alice29.txt", 350), ('I': 401, 'the': 1514, 'Alice': 385, 'was': 352, 'to': 717, 'of': 493, "and': 774, 'she': 497, 'it': 482, 'in': 35) assert equal (commonwords ("alice29.txt", 200), ('I': 401, 'the': 1514, 'Alice': 385, 'Nas': 352, 'to': 717, 'of': 493, 'her': 243, 'and': 774, "she': 497, "it': 4 assert_is_instance (connonwords("alice29.txt", 1000), dict)
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