type() and isinstance() in Python - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check the type of variable x.

Python

x = 10
print(type(x) == [1])

Drag options to blanks, or click blank then click option'

Afloat

Bstr

Clist

Dint

Attempts:

3 left

💡 Hint

Common Mistakes

Using quotes around the type name like 'int' instead of int.

Confusing int with str or float.

The type() function returns the type of the object. Since x is 10, which is an integer, type(x) equals int.

2fill in blank

medium

Complete the code to check if value is an instance of list.

Python

value = [1, 2, 3]
print(isinstance(value, [1]))

Drag options to blanks, or click blank then click option'

Alist

Bdict

Cset

Dtuple

Attempts:

3 left

💡 Hint

Common Mistakes

Using type() instead of isinstance() for type checking.

Choosing the wrong type like tuple or dict.

The isinstance() function checks if the object is an instance of the given type. Here, value is a list, so isinstance(value, list) returns True.

3fill in blank

hard

Fix the error in the code to correctly check if data is a string.

Python

data = 'hello'
if isinstance(data, [1]):
    print('It is a string')

Drag options to blanks, or click blank then click option'

Alist

Bint

Cstr

Dbool

Attempts:

3 left

💡 Hint

Common Mistakes

Using quotes around the type name like 'str'.

Checking against wrong types like int or list.

The isinstance() function needs the type str to check if data is a string. Using str here returns True.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps each word to its length only if the length is greater than 3.

Python

words = ['apple', 'bat', 'car', 'door']
lengths = {word: [1] for word in words if [2] > 3}

Drag options to blanks, or click blank then click option'

Alen(word)

Bword

Attempts:

3 left

💡 Hint

Common Mistakes

Using word instead of len(word) for the value.

Comparing word directly to 3 instead of its length.

The dictionary comprehension maps each word to its length using len(word). The condition filters words with length greater than 3 using len(word) > 3.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase words to their lengths only if the length is greater than 3.

Python

words = ['apple', 'bat', 'car', 'door']
result = { [1]: [2] for word in words if [3] > 3 }

Drag options to blanks, or click blank then click option'

Aword.upper()

Blen(word)

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using word instead of word.upper() for keys.

Comparing word directly to 3 instead of its length.

The dictionary comprehension uses word.upper() as the key, len(word) as the value, and filters words with length greater than 3 using len(word) > 3.

Practice

1. What does the type() function do in Python?

easy

A. It creates a new type.

B. It checks if an object is an instance of a class or its subclass.

C. It converts an object to a string.

D. It returns the exact type of an object.

Solution

1. Step 1: Understand the purpose of type()

   The type() function returns the exact type of the given object, like int, str, or list.

2. Step 2: Compare with other options

   isinstance() checks type inheritance, not exact type. Conversion or creation of types is not done by type().

3. Final Answer:

   It returns the exact type of an object. -> Option D

4. Quick Check:

   type(obj) = exact type [OK]

Hint: Remember: type() gives exact type, isinstance() checks inheritance [OK]

Common Mistakes:

• Confusing type() with isinstance()
• Thinking type() converts or creates types
• Assuming type() checks subclass relationships

2. Which of the following is the correct syntax to check if variable x is an instance of int?

easy

A. type(x) == isinstance(int)

B. isinstance(int, x)

C. isinstance(x, int)

D. type(x) is int()

Solution

1. Step 1: Recall the syntax of isinstance()

   The correct syntax is isinstance(object, class_or_tuple). So, isinstance(x, int) checks if x is an int or subclass.

2. Step 2: Analyze other options

   type(x) == isinstance(int) mixes type() and isinstance() wrongly. isinstance(int, x) reverses arguments. type(x) is int() incorrectly calls int as a function.

3. Final Answer:

   isinstance(x, int) -> Option C

4. Quick Check:

   isinstance(x, int) correct syntax [OK]

Hint: Use isinstance(obj, Type) to check type safely [OK]

Common Mistakes:

• Swapping arguments in isinstance()
• Using type() with wrong comparison
• Calling type names as functions in checks

3. What will be the output of this code?

class Animal:
    pass

class Dog(Animal):
    pass

x = Dog()
print(isinstance(x, Animal))
print(type(x) == Animal)

medium

A. True\nFalse

B. False\nTrue

C. True\nTrue

D. False\nFalse

Solution

1. Step 1: Understand isinstance() with inheritance

   x is an instance of Dog, which is a subclass of Animal. So, isinstance(x, Animal) returns True.

2. Step 2: Understand type() equality check

   type(x) returns Dog, which is not equal to Animal. So, type(x) == Animal returns False.

3. Final Answer:

   True\nFalse -> Option A

4. Quick Check:

   isinstance() checks inheritance, type() exact type [OK]

Hint: isinstance() True if subclass; type() True only if exact type [OK]

Common Mistakes:

• Assuming type() checks inheritance
• Confusing output order
• Thinking isinstance() returns False for subclass

4. Find the error in this code snippet:

def check_number(n):
    if type(n, int):
        return True
    else:
        return False

print(check_number(5))

medium

A. type() is used with two arguments incorrectly.

B. The function should use isinstance() instead of type().

C. The return statements are missing parentheses.

D. The print statement syntax is wrong.

Solution

1. Step 1: Check type() usage

   type() takes one argument to return the type. Using two arguments like type(n, int) is invalid and causes a TypeError.

2. Step 2: Identify correct usage

   To check if n is an int, use isinstance(n, int). The return statements and print syntax are correct.

3. Final Answer:

   type() is used with two arguments incorrectly. -> Option A

4. Quick Check:

   type() needs one argument [OK]

Hint: type() takes one argument; use isinstance() to check type [OK]

Common Mistakes:

• Passing two arguments to type()
• Confusing type() and isinstance() syntax
• Thinking return needs parentheses

5. You have a list of mixed data:

class MyInt(int):
    pass
items = [123, 'hello', 45.6, True, MyInt(456)]

. You want to create a new list containing only integers (including booleans, since they are subclasses of int). Which code correctly uses isinstance() to do this?

hard

A. ints = [x for x in items if type(x) is int or type(x) is bool]

B. ints = [x for x in items if isinstance(x, int)]

C. ints = [x for x in items if isinstance(x, bool)]

D. ints = [x for x in items if type(x) == int]

Solution

1. Step 1: Understand subclass behavior with isinstance()

   Booleans are subclasses of int in Python, so isinstance(True, int) returns True. Using isinstance(x, int) includes both ints and bools.

2. Step 2: Compare with other options

   ints = [x for x in items if type(x) == int] excludes bools because type(True) is bool, not int. ints = [x for x in items if isinstance(x, bool)] selects only bools, not ints. ints = [x for x in items if type(x) is int or type(x) is bool] is verbose and misses subclass flexibility.

3. Final Answer:

   ints = [x for x in items if isinstance(x, int)] -> Option B

4. Quick Check:

   isinstance() includes subclasses [OK]

Hint: Use isinstance(x, int) to include bools as ints [OK]

Common Mistakes:

• Using type() == int excludes bools
• Selecting only bools instead of ints
• Using complex or incorrect type checks