Question 1 : Write a Program to rename a key from the dictionary.
Given Dictionary:
employee_detail = {
"name" : "Tim",
"age" : 25,
"gender" : "Male"
}Expected Output:
{'age': 25, 'gender': 'Male', 'first_name': 'Tim'}[bg_collapse view=”button-green” color=”#ffffff” icon=”eye” expand_text=”Show Answer” collapse_text=”Hide Answer”]
Solution:
employee_detail = {
"name" : "Tim",
"age" : 25,
"gender" : "Male"
}
employee_detail["first_name"] = employee_detail.pop("name")
print(employee_detail)Output:
{'age': 25, 'gender': 'Male', 'first_name': 'Tim'}[/bg_collapse]
Question 2 : Write a Program to remove a list of keys from the dictionary.
Given Dictionary:
marks = {
"Bill" : 78,
"Tim" : 75,
"Steve" : 80
}
to_remove = ["Tim", "Jeff"]Expected Output:
{'Bill': 78, 'Steve': 80}[bg_collapse view=”button-green” color=”#ffffff” icon=”eye” expand_text=”Show Answer” collapse_text=”Hide Answer”]
Solution:
We can achieve this task using the two approaches:
Approach 1: Using the for loop
marks = {
"Bill" : 78,
"Tim" : 75,
"Steve" : 80,
"Jeff" : 79
}
to_remove = ["Tim", "Jeff"]
for name in to_remove:
if name in marks:
marks.pop(name)
print(marks)Approach 2: Using comprehension
marks = {
"Bill" : 78,
"Tim" : 75,
"Steve" : 80,
"Jeff" : 79
}
to_remove = ["Tim", "Jeff"]
new_marks = {name:marks[name] for name in marks.keys() - to_remove}# in to_remove:
print(new_marks)Output:
{'Bill': 78, 'Steve': 80}[/bg_collapse]
Question 3 : How can we find maximum value from the dictionary?
Given Dictionary:
marks = {
"Bill" : 78,
"Tim" : 75,
"Steve" : 80
}Expected Output:
Steve[bg_collapse view=”button-green” color=”#ffffff” icon=”eye” expand_text=”Show Answer” collapse_text=”Hide Answer”]
Solution:
Using the max() function.
marks = {
"Bill" : 78,
"Tim" : 75,
"Steve" : 80
}
print(max(marks, key = marks.get))Output:
Steve[/bg_collapse]
Question 4 : Create a new dictionary with specific keys from the old one
Given Dictionary:
marks = {
"Bill" : 78,
"Tim" : 75,
"Steve" : 80,
"Jeff" : 79
}Expected Output:
{'Tim': 75, 'Jeff': 79}[bg_collapse view=”button-green” color=”#ffffff” icon=”eye” expand_text=”Show Answer” collapse_text=”Hide Answer”]
Solution:
Approach 1: Using a for loop and a temporary variable.
marks = {
"Bill" : 78,
"Tim" : 75,
"Steve" : 80,
"Jeff" : 79
}
new_keys = ["Tim", "Jeff"]
new_dict = {}
for name in new_keys:
new_dict[name] = marks[name]
print(new_dict)Approach 2: Using dictionary comprehension.
marks = {
"Bill" : 78,
"Tim" : 75,
"Steve" : 80,
"Jeff" : 79
}
new_keys = ["Tim", "Jeff"]
new_dict = {name: marks[name] for name in new_keys}
print(new_dict)Approach 3: Using the update method.
marks = {
"Bill" : 78,
"Tim" : 75,
"Steve" : 80,
"Jeff" : 79
}
new_keys = ["Tim", "Jeff"]
new_dict = {}
for name in new_keys:
new_dict.update({name: marks[name]})
print(new_dict)Output:
Steve[/bg_collapse]
Question 5 : Check if the value is present in the dictionary.
Given Dictionary:
marks = {"Bill" : 87, "Steve" : 85, "Jeff" : 81}Value to check:
81[bg_collapse view=”button-green” color=”#ffffff” icon=”eye” expand_text=”Show Answer” collapse_text=”Hide Answer”]
Solution:
Approach 1: Using values() method.
marks = {"Bill" : 87, "Steve" : 85, "Jeff" : 81}
value = 81
if value in marks.values():
print("Value is Present")
else:
print("Value is not Present")Approach 2: Using for loop.
marks = {"Bill" : 87, "Steve" : 85, "Jeff" : 81}
value = 81
for i in marks:
if value == marks[i]:
print("Value is Present")
break;
else:
print("Value is not presnt")Output:
Value is Present[/bg_collapse]
Question 6 : We have 3 dictionaries of marks of marks. How can we create a single dictionary by nesting three dictionaries in the one dictionary.
Given Dictionaries:
bill = {1 : 82, 2 : 84, 3 : 86}
steve = {1 : 81, 2 : 82, 3 : 89}
jeff = {1 : 84, 2 : 85, 3 : 88}Expected Output:
{'bill': {1: 82, 2: 84, 3: 86}, 'steve': {1: 81, 2: 82, 3: 89}, 'jeff': {1: 84, 2: 85, 3: 88}}[bg_collapse view=”button-green” color=”#ffffff” icon=”eye” expand_text=”Show Answer” collapse_text=”Hide Answer”]
Solution:
bill = {1 : 82, 2 : 84, 3 : 86}
steve = {1 : 81, 2 : 82, 3 : 89}
jeff = {1 : 84, 2 : 85, 3 : 88}
marks = {"bill" : bill, "steve" : steve, "jeff" : jeff}
print(marks)Output:
{'bill': {1: 82, 2: 84, 3: 86}, 'steve': {1: 81, 2: 82, 3: 89}, 'jeff': {1: 84, 2: 85, 3: 88}}[/bg_collapse]
Question 7 : We have 3 dictionaries of marks. How can we create a single dictionary by nesting three dictionaries in the one dictionary?
Given Dictionaries:
bill = {1 : 82, 2 : 84, 3 : 86}
steve = {1 : 81, 2 : 82, 3 : 89}
jeff = {1 : 84, 2 : 85, 3 : 88}Expected Output:
{'bill': {1: 82, 2: 84, 3: 86}, 'steve': {1: 81, 2: 82, 3: 89}, 'jeff': {1: 84, 2: 85, 3: 88}}[bg_collapse view=”button-green” color=”#ffffff” icon=”eye” expand_text=”Show Answer” collapse_text=”Hide Answer”]
Solution:
bill = {1 : 82, 2 : 84, 3 : 86}
steve = {1 : 81, 2 : 82, 3 : 89}
jeff = {1 : 84, 2 : 85, 3 : 88}
marks = {"bill" : bill, "steve" : steve, "jeff" : jeff}
print(marks)Output:
{'bill': {1: 82, 2: 84, 3: 86}, 'steve': {1: 81, 2: 82, 3: 89}, 'jeff': {1: 84, 2: 85, 3: 88}}[/bg_collapse]
Question 9 : How can we check if the dictionary is empty?
[bg_collapse view=”button-green” color=”#ffffff” icon=”eye” expand_text=”Show Answer” collapse_text=”Hide Answer”]
Solution:
dict1 = {}
dict2 = {1:"one", 2:"two"}
if dict1 == dict():
print("Dictionary dict1 is empty.")
if dict2 == dict():
print("Dictionary dict2 is empty.")Output:
Dictionary dict1 is empty.[/bg_collapse]
Question 9 : We have a dictionary with 9 votes for two candidates, ‘a’ and ‘b’. How can we count the number of votes each candidate has got and the winner?
Given Dictionary:
v = {1:'a', 2:'b', 3:'a', 4:'a', 5:'b', 6:'a', 7:'a', 8:'b', 9:'a'}[bg_collapse view=”button-green” color=”#ffffff” icon=”eye” expand_text=”Show Answer” collapse_text=”Hide Answer”]
Solution:
v = {1:'a', 2:'b', 3:'a', 4:'a', 5:'b', 6:'a', 7:'a', 8:'b', 9:'a'}
count = {}
for i in v:
if v[i] in count:
count[v[i]] += 1
else:
count[v[i]] = 1
print("Final result =",count)
print("Winner =", max(count, key = count.get)) Output:
Final result = {'a': 6, 'b': 3}
Winner = a[/bg_collapse]