Complete the following Python CODE. Only modify this file. Put code in after the
ID: 3745335 • Letter: C
Question
Complete the following Python CODE. Only modify this file. Put code in after the line ######### YOUR CODE STARTS HERE ################ #### Note: In the case of dictionaries, use the keys as the determining factor for all work on the data structures #### Note: You can give preference to data structure a over b, but no use case will require that you have done this. #### This means if a is a dictionary with keys "a","b", and "c" and b is a dictionary with key of "a" #### You will not be tested as two which value for key "a" would be in the result #### Note: You will only be tested in which both data structures are the same type, no a is a list and b is a tuple. #### The result of the methods will be tested to ensure they are the same data structure as a and b def unionOfTwoDataStructures(a,b): result = None #Find the concatenation of the data structures (with no uniqueness guarantee) a and b and set the return variable 'result' to this value #Note: if you ensure that each value from a and b only occurs once, it may make other methods easier to complete #Note: While using concatenation will get you most of the way to this, it will not result in only unique values if type(a) is list and type(b) is list: ######### YOUR CODE STARTS HERE ################ ######### YOUR CODE ENDS HERE ################ elif type(a) is tuple and type(b) is tuple: ######### YOUR CODE STARTS HERE ################ ######### YOUR CODE ENDS HERE ################ elif type(a) is dict and type(b) is dict: ######### YOUR CODE STARTS HERE ################ ######### YOUR CODE ENDS HERE ################ else: print "a and b have to both be either lists, tuples, or dictionaries" return result def intersectionOfTwoDataStructures(a,b): result = None #Find the elements of both of the data structures a and b and set the return variable 'result' to this value #Note: if you ensure that each value from a and b only occurs once, it may make other methods easier to complete if type(a) is list and type(b) is list: ######### YOUR CODE STARTS HERE ################ ######### YOUR CODE ENDS HERE ################ elif type(a) is tuple and type(b) is tuple: ######### YOUR CODE STARTS HERE ################ ######### YOUR CODE ENDS HERE ################ elif type(a) is dict and type(b) is dict: ######### YOUR CODE STARTS HERE ################ ######### YOUR CODE ENDS HERE ################ else: print "a and b have to both be either lists, tuples, or dictionaries" return result def symmetricDifferenceOfTwoDataStructures(a,b): result = None #Find the elements that are not common to both data structures a and b and set the return variable 'result' to this value #Note: you may be able to use other methods together to get this done without much extra code, look up venn diagrams to see # how this could happen ######### YOUR CODE STARTS HERE ################ if type(a) is list and type(b) is list: ######### YOUR CODE STARTS HERE ################ ######### YOUR CODE ENDS HERE ################ elif type(a) is tuple and type(b) is tuple: ######### YOUR CODE STARTS HERE ################ ######### YOUR CODE ENDS HERE ################ elif type(a) is dict and type(b) is dict: ######### YOUR CODE STARTS HERE ################ ######### YOUR CODE ENDS HERE ################ else: print "a and b have to both be either lists, tuples, or dictionaries" ######### YOUR CODE ENDS HERE ################ return result def equalityOfTwoDataStructures(a,b): result = None #Determine if the two data structures passed in as variables are equivalent and place the associated # boolean value of True or False into the variable result ######### YOUR CODE STARTS HERE ################ if type(a) is list and type(b) is list: #Note: Equality of two lists means that they contain all of the same values, it does not mean # mean that both lists have them in same order or do not include duplicates ######### YOUR CODE STARTS HERE ################ ######### YOUR CODE ENDS HERE ################ elif type(a) is tuple and type(b) is tuple: ######### YOUR CODE STARTS HERE ################ ######### YOUR CODE ENDS HERE ################ elif type(a) is dict and type(b) is dict: ######### YOUR CODE STARTS HERE ################ ######### YOUR CODE ENDS HERE ################ else: print "a and b have to both be either lists, tuples, or dictionaries" ######### YOUR CODE ENDS HERE ################ return result def test_all_methods(a,b): result = [] result.append(unionOfTwoDataStructures(a,b)) result.append(intersectionOfTwoDataStructures(a,b)) result.append(symmetricDifferenceOfTwoDataStructures(a,b)) result.append(equalityOfTwoDataStructures(a,b)) return result
Explanation / Answer
Python:
#### Note: In the case of dictionaries, use the keys as the determining factor for all work on the data structures
#### Note: You can give preference to data structure a over b, but no use case will require that you have done this.
#### This means if a is a dictionary with keys "a","b", and "c" and b is a dictionary with key of "a"
#### You will not be tested as two which value for key "a" would be in the result
#### Note: You will only be tested in which both data structures are the same type, no a is a list and b is a tuple.
#### The result of the methods will be tested to ensure they are the same data structure as a and b
def unionOfTwoDataStructures(a,b):
result = None
#Find the concatenation of the data structures (with no uniqueness guarantee) a and b and set the return variable 'result' to this value
#Note: if you ensure that each value from a and b only occurs once, it may make other methods easier to complete
#Note: While using concatenation will get you most of the way to this, it will not result in only unique values
if type(a) is list and type(b) is list:
######### YOUR CODE STARTS HERE ################
result=[]
for i in a:
result.append(i)
for i in b:
result.append(i)
######### YOUR CODE ENDS HERE ################
elif type(a) is tuple and type(b) is tuple:
######### YOUR CODE STARTS HERE ################
result=[]
for i in list(a):
result.append(i)
for i in list(b):
result.append(i)
result=tuple(result)
######### YOUR CODE ENDS HERE ################
elif type(a) is dict and type(b) is dict:
######### YOUR CODE STARTS HERE ################
result={}
for i in a:
result[i]=a[i]
for i in b:
if(i not in result): ## I am giving priority to a over b so if element is in a then consider it as prior.
result[i]=b[i]
######### YOUR CODE ENDS HERE ################
else:
print("a and b have to both be either lists, tuples, or dictionaries")
return result
def intersectionOfTwoDataStructures(a,b):
result = None
#Find the elements of both of the data structures a and b and set the return variable 'result' to this value
#Note: if you ensure that each value from a and b only occurs once, it may make other methods easier to complete
if type(a) is list and type(b) is list:
######### YOUR CODE STARTS HERE ################
result=[]
for i in a:
if(i in b):
result.append(i)
for i in b:
if(i in a and i not in b):
result.append(i)
######### YOUR CODE ENDS HERE ################
elif type(a) is tuple and type(b) is tuple:
######### YOUR CODE STARTS HERE ################
result=[]
for i in list(a):
if(i in b):
result.append(i)
for i in list(b):
if(i in a and i not in b):
result.append(i)
result=tuple(result)
######### YOUR CODE ENDS HERE ################
elif type(a) is dict and type(b) is dict:
######### YOUR CODE STARTS HERE ################
result={}
for i in a:
if(i in b):
result[i]=a[i] ## Here i am again giving priority to a over b.
######### YOUR CODE ENDS HERE ################
else:
print("a and b have to both be either lists, tuples, or dictionaries")
return result
def symmetricDifferenceOfTwoDataStructures(a,b):
result = None
#Find the elements that are not common to both data structures a and b and set the return variable 'result' to this value
#Note: you may be able to use other methods together to get this done without much extra code, look up venn diagrams to see
# how this could happen
######### YOUR CODE STARTS HERE ################
if type(a) is list and type(b) is list:
######### YOUR CODE STARTS HERE ################
result=[]
for i in a:
if(i not in b):
result.append(i)
for i in b:
if(i not in a):
result.append(i)
######### YOUR CODE ENDS HERE ################
elif type(a) is tuple and type(b) is tuple:
######### YOUR CODE STARTS HERE ################
result=[]
for i in list(a):
if(i not in b):
result.append(i)
for i in list(b):
if(i not in a):
result.append(i)
result=tuple(result)
######### YOUR CODE ENDS HERE ################
elif type(a) is dict and type(b) is dict:
######### YOUR CODE STARTS HERE ################
result={}
for i in a:
if(i not in b):
result[i]=a[i]
for i in b:
if(i not in a):
result[i]=b[i]
######### YOUR CODE ENDS HERE ################
else:
print("a and b have to both be either lists, tuples, or dictionaries")
######### YOUR CODE ENDS HERE ################
return result
def equalityOfTwoDataStructures(a,b):
result = None
#Determine if the two data structures passed in as variables are equivalent and place the associated
# boolean value of True or False into the variable result
######### YOUR CODE STARTS HERE ################
if type(a) is list and type(b) is list:
result=True
for i in a:
if(i not in b):
result=False
for i in b:
if(i not in a):
result=False
#Note: Equality of two lists means that they contain all of the same values, it does not mean
# mean that both lists have them in same order or do not include duplicates
######### YOUR CODE STARTS HERE ################
######### YOUR CODE ENDS HERE ################
elif type(a) is tuple and type(b) is tuple:
######### YOUR CODE STARTS HERE ################
result=True
for i in list(a):
if(i not in b):
result=False
for i in list(b):
if(i not in a):
result=False
######### YOUR CODE ENDS HERE ################
elif type(a) is dict and type(b) is dict:
######### YOUR CODE STARTS HERE ################
result=True
for i in a:
if(i not in b):
result=False
for i in b:
if(i not in a):
result=False
######### YOUR CODE ENDS HERE ################
else:
print("a and b have to both be either lists, tuples, or dictionaries")
######### YOUR CODE ENDS HERE ################
return result
def test_all_methods(a,b):
result = []
result.append(unionOfTwoDataStructures(a,b))
result.append(intersectionOfTwoDataStructures(a,b))
result.append(symmetricDifferenceOfTwoDataStructures(a,b))
result.append(equalityOfTwoDataStructures(a,b))
return result