Question

Hey guys I need help with this question with 3 sub-problems.

f test remove short synonyms () Define the remove shorti synonyms function which is passed a dictionary as a parameter- The keys of the parameter dictionary are words and the corresponding values are 1ists of synonyms (synonyms are words which have the same or nearly the same meaning). The function romoves all the eynonyme which have ous than 8 charactors from each corresponding list of synonyms-As well, the funet ion sorts each corresponding list of synonyms- For example, the following code: synonyms dict('look : 'gaze see, 'glance, 'watch', 'peruse' 1, 'put': I'place, 'set, attach , keep', 'save', 'set aside', effect', achieve', "dobuild' 1 beautiful l 'prettyovely handsome' 'dazzling', 'splendid', magnificent'] slow: I 'unhurried, 'gradual'leisurely', 'late, 'behind', tediousslack 1, dangerous'periloushazardous, 'uncertain' remove short_synonyms ( synonyms dict) print("1) print dict in key ordertsynonyms dict) printo synonyms dict 'come ('approach 'advance' near, 'arrive, reach show E'displayexhibit, 'present, note, 'point to'. indicate 'explain reveal', prove demonstrate', 'expose'1, good'excellent fine superior wonderful', 'grand', superb edifying 1 bad: 'evil',immoral wicked rotten'contaminated, spoiled'. defectivesubstandard, faulty improper, inappropriatel remove short synonyms (synonyms dict) print( "2.") print dict in key order(synonyms dict) prints: dangerous : [ hazardous' perilous'uncertain' ] look : ( put : ['set aside' 1 slow s ['leisurely, 'unhurried' l bad : [ 'contaminated', 'defective', improper, 'inappropriate ' substandard' 1 come : approach 1 good : I'edifying, excellent', 'superior', 'wonderful'l show [ demonstrate', 'indicate', point tol def remove short_synonyms (synonyms dict): pass # 6666666666666666666666666666666666666666666666666 # contains-keys-and values() fine the contains keys and values) function which is passed two dict objects as parameters, dicti and dict2. The two parameter ries both have corresponding values which are lists of elements (numbers or strings). The function return True if the following conditions are met: . dictl contains all the keys which are in dict2 (dictl may contain extra keys), and, . the elements in all the corresponding value lists of diet2 are also elements in one or more of the corresponding value lists of dictl. Note: when testing this part of the condition, the elements can be in any order and in any of the corresponding value lists, e.g., if one of the corresponding values lists of dict2 is [4, 2] and any one of the corresponding value lists of dictl contains the element 4 and any one of the corresponding value lists of dicti contains the element 2, this part of the condition is satisfied. The function returns False in al1 other cases. For example, the following code: dictl 0 dict2 = {} print( "1." contains keys and values (dicti, dict2)) dicti f'a: [4, 3 d 16, 21,"z": "t":[2, 231) dict2 [2, 3, 6, 23 a: 141) print("2., contains keys and values (dicti, dict2)) dict2t"a" (3, 6, 3, "p [6, 11) print ("3., contains_keys and values(dictl, dict2)) dicti "a": (6, 3, "p": [1) print("4.", contains_keys and values(dictl, dict2)) dictl ("a": (6, 3, "p": 'a'l, "t": I's'1 dict2 = {"a": [3, 6, 31, "p": ['a, ] , "s". I'ah print ("5., contains keys and values(dictl, dict2)) prints: 1. True 2. True 3. False 4. False 5. False def contains_keys_ and_values (filename) pass + get_previous words dict) Define the get previous words dict () function which is passed a string of text as a parameter. The function returns a dictionary with keys which are all the unique words from the text, and, the corresponding values which are lists of all the unique words in the text which come before the key word. Note that in each corresponding list of previous words, the same word should not appear more than once. Notes: the first word in the sentence will initially have the empty string as its previous word, , you can assume that the text is all in lower case and contains no punctuation , the testing code makes use of the print dict_in key order (a dict) which prints the dictionary pairs in sorted key order. , each list of previous words must be sorted into ascending order. For example, the following code: sentence 'a man we saw saw a saw previous words dict get previous_ words dict (sentence) print dict in key order(previous_words_dïet) print() sentence, my favourite painting is the painting i did of my dog in that previous-words-dict get-previous-words-dict (sentence) print dict in key order(previous_words_dict) prints: man : ['a saw : t'a, saw', 'we'1 we: Iman dog : [my 1 favourite : [my'1 í : ['painting') n : ['dog', painting] is : [ painting' 1 of : I'did'] painting : ['favourite', 'that' 'the'l the : ['is def get previous words dict (text) return t)

Answer 1

NOTE: I Hope the below solution meets your requirements as per the Question. Please Do UPVOTE if it Does. :)

All test cases of the question are being checked and shown in below screenshots.

## Python Code:

## Part 1 of the problem
print()
print('-'*10,'PART 1 OF THE PROBLEM','-'*10)
print()

## Function to remove the short synonyms
def remove_short_synonyms(synonyms_dict):
# Iterating each key,value pair in the dictionary
for key,value in synonyms_dict.items():
# Empty list to store the new values for the key
new_value = []
# checking the lenght of the words in the value
for word in value:
if len(word) >= 8:
new_value.append(word)
synonyms_dict[key] = new_value
# returning the dictionary
return synonyms_dict

## Function to print the dictionary in sorted order
def print_dict_in_key_order(synonyms_dict):
for key,value in sorted(synonyms_dict.items()):
print(key,':',sorted(value))

# Storing the words with their synonyms in the form of dictionary
synonyms_dict = {'look' : ['gaze', 'see', 'glance', 'watch', 'persue'],
'put' : ['place', 'set', 'attach', 'keep', 'save', 'set aside', 'achieve', 'do', 'build'],
'beautiful' : ['pretty', 'lovely', 'handsome', 'dazzling', 'splendid', 'magnificient'],
'slow' : ['unhurried', 'gradual', 'leisurely', 'late', 'behind', 'tedious', 'slack'],
'dangerous' : ['perilous', 'hazardous', 'uncertain']
}

# Removing the short synonyms from the dictionary
synonyms_dict = remove_short_synonyms(synonyms_dict)
print('1.')
# printing the dictionary
print_dict_in_key_order(synonyms_dict)
print()

# Storing the words with their synonyms in the form of dictionary
synonyms_dict = {'come' : ['approach', 'advance', 'near', 'arrive', 'reach'],
'show' : ['display', 'exhibit', 'present', 'note', 'point to', 'indicate', 'explain', 'reveal', 'prove', 'demonstrate', 'expose'],
'good' : ['excellent', 'fine', 'superior', 'wonderful', 'grand', 'superb', 'edifying'],
'bad' : ['evil', 'immoral', 'wicked', 'rotten', 'contaminated', 'spoiled', 'defective', 'substandard', 'faulty', 'improper', 'inappropriate']
}

# Removing the short synonyms from the dictionary
synonyms_dict = remove_short_synonyms(synonyms_dict)
print('2.')
# printing the dictionary
print_dict_in_key_order(synonyms_dict)
print()
print()

## Part 2 of the problem
print()
print('-'*10,'PART 2 OF THE PROBLEM','-'*10)
print()

# Function to check the donditions of dict1 and dict2
def contains_keys_and_values(dict1, dict2):
# If any of the condition fails return False else return True
# checking if dict1 contains all the keys of dict2
for key in dict2.keys():
if key not in dict1.keys():
return False
dict1_values = []
# checking if the values of dict2 is in any corresponding value list in dict1
for value in dict1.values():
dict1_values.extend(value)
for value in dict2.values():
for ele in value:
if ele not in dict1_values:
return False
# if both the condition is satisfied return True
return True

# Checking the working of the contains_keys_and_values function for different set of values
dict1 = {}
dict2 = {}
print('1.', contains_keys_and_values(dict1, dict2))
dict1 = {'a': [4,3] , 'd' : [6,2], 'z':[], 't': [2,23]}
dict2 = {'z': [2,3,6,23], 'a':[4]}
print('2.', contains_keys_and_values(dict1, dict2))
dict1 = {'a': [6,3], 'p':[]}
dict2 = {'a': [3,6,3], 'p':[6,1]}
print('3.', contains_keys_and_values(dict1, dict2))
dict1 = {'a': [6,3], 'p': []}
dict2 = {'a': [3,6,3], 'p': ['a']}
print('4.', contains_keys_and_values(dict1, dict2))
dict1 = {'a': [6,3], 'p': ['a'], 't': ['s']}
dict2 = {'a': [3,6,3], 'p':['a'], 's': ['a']}
print('5.', contains_keys_and_values(dict1, dict2))

## Part 3 of the problem
print()
print('-'*10,'PART 3 OF THE PROBLEM','-'*10)
print()

# Function to get the previous words of the dictionary
def get_previous_words_dict(text):
# converting the text into list
text = text.split()
# creating an empty dictionary to store the unique words with value as previous words
previous_words_dict = {}
# getting the unique words from the text in the sorted order
keys = sorted(set(text))
# iterating through each key
for key in keys:
# Empty list to store the previous words
previous_words = []
# finding the prevous words for the key
for index in range(len(text)):
if key == text[index]:
if index == 0:
previous_words.append('')
else:
previous_words.append(text[index-1])
# adding the key with previous words list to the dictionary
previous_words_dict[key] = previous_words
# returning the dictinoary
return previous_words_dict

# Checking the working of get_previous_words_dict function for different values
sentence = 'a man we saw saw a saw'
previous_words_dict = get_previous_words_dict(sentence)
print_dict_in_key_order(previous_words_dict)
print()

sentence = 'my favourite painting is the painting i did of my dog in that painting in my den'
previous_words_dict = get_previous_words_dict(sentence)
print_dict_in_key_order(previous_words_dict)

## Screenshots of the Code in IDLE:

## Screenshots of the Output: