Local scope in Python - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define a local variable x inside the function.

Python

def my_function():
    [1] = 10
    print(x)

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

Ax

Bz

Cy

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different variable name than the one printed.

Defining the variable outside the function.

The variable x is defined locally inside the function. This means it only exists inside my_function.

2fill in blank

medium

Complete the code to return the local variable result from the function.

Python

def calculate():
    result = 5 + 3
    return [1]

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

Avalue

Bresult

Ctotal

Dx

Attempts:

3 left

💡 Hint

Common Mistakes

Returning a variable not defined inside the function.

Returning a variable with a different name.

The function returns the local variable result which holds the sum.

3fill in blank

hard

Fix the error by completing the code to access the local variable count inside the function.

Python

def counter():
    count = 0
    count += 1
    print([1])

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

Avalue

Bcounter

Ctotal

Dcount

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to print a variable not defined inside the function.

Using the function name instead of the variable.

The variable count is local and must be used inside the function to print its value.

4fill in blank

hard

Fill both blanks to create a dictionary with keys as words and values as their lengths, only for words longer than 3 letters.

Python

words = ['apple', 'cat', 'banana', 'dog']
lengths = {word: [1] for word in words if len(word) [2] 3}

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

Alen(word)

B>

C<

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself as the value instead of its length.

Using the wrong comparison operator.

The dictionary comprehension uses len(word) for values and filters words with length greater than 3.

5fill in blank

hard

Fill all three blanks to create a dictionary with uppercase keys and values only if the value is greater than 0.

Python

data = {'a': 1, 'b': 0, 'c': 3}
result = { [1]: [2] for k, v in data.items() if v [3] 0 }

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

Ak.upper()

Bv

C>

Dk

Attempts:

3 left

💡 Hint

Common Mistakes

Not converting keys to uppercase.

Including values that are zero or less.

Swapping keys and values.

The dictionary comprehension uses uppercase keys, original values, and filters values greater than zero.

Practice

1. What does local scope mean in Python?

easy

A. Variables can be used anywhere in the program

B. Variables exist only inside the function where they are created

C. Variables are shared between all functions

D. Variables are stored permanently on the disk

Solution

1. Step 1: Understand the meaning of local scope

   Local scope means a variable is created inside a function and only exists there.

2. Step 2: Compare options with this meaning

   Only Variables exist only inside the function where they are created correctly states that variables exist only inside their function.

3. Final Answer:

   Variables exist only inside the function where they are created -> Option B

4. Quick Check:

   Local scope = variables inside function only [OK]

Hint: Local variables live only inside their function [OK]

Common Mistakes:

• Thinking local variables can be used outside the function
• Confusing local scope with global scope
• Believing variables are shared across functions

2. Which of the following is the correct way to define a local variable inside a function?

easy

A. def func():
x = 5

B. x = 5
def func():
print(x)

C. def func():
global x
x = 5

D. def func():
return x

Solution

1. Step 1: Identify local variable definition

   A local variable is created by assigning a value inside a function without global keyword.

2. Step 2: Check each option

   def func(): x = 5 assigns x inside the function, making it local. Others either use global or no assignment inside function.

3. Final Answer:

   def func():\n x = 5 -> Option A

4. Quick Check:

   Assign inside function = local variable [OK]

Hint: Assign variable inside function without global for local [OK]

Common Mistakes:

• Using global keyword when not needed
• Assigning variable outside function expecting it local
• Trying to return variable not defined inside function

3. What will be the output of this code?

def greet():
    message = "Hello"
    print(message)

greet()
print(message)

medium

A. Hello
NameError

B. Hello
Hello

C. NameError
Hello

D. NameError
NameError

Solution

1. Step 1: Understand variable scope in the code

   Variable 'message' is defined inside greet(), so it is local to that function.

2. Step 2: Trace the print statements

   Calling greet() prints 'Hello'. Then print(message) outside function causes NameError because 'message' is not defined globally.

3. Final Answer:

   Hello\nNameError -> Option A

4. Quick Check:

   Local variable outside function causes NameError [OK]

Hint: Local variables can't be printed outside their function [OK]

Common Mistakes:

• Assuming local variable is accessible globally
• Expecting both prints to show 'Hello'
• Ignoring NameError on second print

4. Find the error in this code and fix it:

def add():
    result = a + b
    print(result)

add()

Assuming a = 2 and b = 3 are defined outside the function.

medium

A. Add 'global a, b' inside add()

B. No error, code runs fine

C. Define a and b inside add()

D. Pass a and b as parameters to add()

Solution

1. Step 1: Identify variable scope issue

   a and b are defined outside but used inside add() without global or parameters, causing NameError.

2. Step 2: Fix by passing variables as parameters

   Passing a and b as parameters to add() allows access without global keyword.

3. Final Answer:

   Pass a and b as parameters to add() -> Option D

4. Quick Check:

   Use parameters to access outside variables inside function [OK]

Hint: Pass outside variables as parameters to use inside function [OK]

Common Mistakes:

• Using global keyword unnecessarily
• Defining variables inside function losing outside values
• Ignoring NameError from missing variables

5. How can you modify this code to keep track of how many times count_calls() is called, using local scope only?

def count_calls():
    calls = 0
    calls += 1
    print(f"Called {calls} times")

count_calls()
count_calls()

hard

A. Declare calls as global variable

B. Define calls outside function and modify inside

C. Use a default argument to store calls count

D. Use a class to store calls count

Solution

1. Step 1: Understand why calls resets

   Variable calls is local and resets to 0 each call, so count never increases.

2. Step 2: Use default argument to keep state locally

   Using a default argument like calls=[0] keeps count inside function without global or external variables.

3. Final Answer:

   Use a default argument to store calls count -> Option C

4. Quick Check:

   Default argument keeps local state across calls [OK]

Hint: Use default mutable argument to keep count inside function [OK]

Common Mistakes:

• Using global variable instead of local trick
• Defining calls outside function losing local scope
• Not realizing local variables reset each call