Dictionary comprehension with condition in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of dictionary comprehension with a simple condition

What is the output of this Python code?

Python

result = {x: x**2 for x in range(5) if x % 2 == 0}
print(result)

A {0: 0, 1: 1, 2: 4, 3: 9, 4: 16}

B {1: 1, 3: 9}

C {0: 0, 2: 4, 4: 16}

D {}

💡 Hint

Remember, the condition filters keys where x is even.

The comprehension includes only keys where x % 2 == 0, so only 0, 2, and 4 are included with their squares.

❓ Predict Output

intermediate

Dictionary comprehension with else in value expression

What does this code print?

Python

result = {x: ('even' if x % 2 == 0 else 'odd') for x in range(3)}
print(result)

A {0: 'even', 1: 'odd', 2: 'even'}

B {0: 'odd', 1: 'even', 2: 'odd'}

C {0: 'even', 2: 'even'}

D {1: 'odd'}

💡 Hint

The value uses a conditional expression to label even or odd.

Each key from 0 to 2 is included. Values are 'even' if key is even, else 'odd'.

❓ Predict Output

advanced

Output of nested dictionary comprehension with condition

What is the output of this code?

Python

result = {x: {y: x*y for y in range(3) if y != x} for x in range(3)}
print(result)

A {0: {}, 1: {}, 2: {}}

B {0: {1: 1, 2: 2}, 1: {0: 0, 2: 2}, 2: {0: 0, 1: 1}}

C {0: {0: 0, 1: 0, 2: 0}, 1: {0: 0, 1: 1, 2: 2}, 2: {0: 0, 1: 2, 2: 4}}

D {0: {1: 0, 2: 0}, 1: {0: 0, 2: 2}, 2: {0: 0, 1: 2}}

💡 Hint

Inner comprehension excludes y equal to x.

For each x, inner dict includes y from 0 to 2 except y == x, mapping y to x*y.

❓ Predict Output

advanced

Dictionary comprehension with complex condition and filtering

What will be printed by this code?

Python

data = {'a': 1, 'b': 2, 'c': 3, 'd': 4}
result = {k: v for k, v in data.items() if v % 2 == 0 and k != 'd'}
print(result)

A {'a': 1, 'c': 3}

B {'b': 2}

C {'b': 2, 'd': 4}

D {}

💡 Hint

Filter keys with even values but exclude key 'd'.

Only 'b' has an even value and key not equal to 'd', so only {'b': 2} remains.

🧠 Conceptual

expert

Count of items in dictionary comprehension with multiple conditions

How many items are in the dictionary created by this comprehension? {num: num**3 for num in range(10) if num % 2 == 1 if num > 5}

A 2

B 5

C 4

D 3

💡 Hint

Count numbers from 0 to 9 that are odd and greater than 5.

Numbers 7 and 9 satisfy both conditions, so the dictionary has 2 items.

Practice

1.

What does the following dictionary comprehension do?
{k: v for k, v in {'a': 1, 'b': 2, 'c': 3}.items() if v > 1}

easy

A. {'a': 1, 'b': 2, 'c': 3} - keeps all items

B. {'a': 1} - keeps items with values equal to 1

C. {'b': 2, 'c': 3} - keeps items with values greater than 1

D. {} - removes all items

Solution

1. Step 1: Understand the dictionary comprehension structure

   The comprehension loops over each key-value pair in the original dictionary.

2. Step 2: Apply the condition if v > 1

   Only pairs where the value is greater than 1 are included in the new dictionary.

3. Final Answer:

   {'b': 2, 'c': 3} - keeps items with values greater than 1 -> Option C

4. Quick Check:

   Filter values > 1 = {'b': 2, 'c': 3} [OK]

Hint: Look for the condition after the for loop to filter items [OK]

Common Mistakes:

• Ignoring the condition and including all items
• Confusing keys and values in the condition
• Using wrong comparison operator

2.

Which of the following is the correct syntax for a dictionary comprehension with a condition?

?

easy

A. {k: v for k, v in d.items() if v % 2 == 0}

B. {k: v if v % 2 == 0 for k, v in d.items()}

C. {k: v for k, v in d.items() where v % 2 == 0}

D. {k: v for k, v in d.items() when v % 2 == 0}

Solution

1. Step 1: Recall dictionary comprehension syntax

   The correct syntax is {key: value for key, value in iterable if condition}.

2. Step 2: Identify the correct option

   {k: v for k, v in d.items() if v % 2 == 0} matches the correct syntax with if after the loop. Others use invalid keywords or wrong order.

3. Final Answer:

   {k: v for k, v in d.items() if v % 2 == 0} -> Option A

4. Quick Check:

   Correct syntax uses 'if' after for loop [OK]

Hint: Remember: condition goes after the for clause in comprehension [OK]

Common Mistakes:

• Placing 'if' before the for loop
• Using 'where' or 'when' instead of 'if'
• Incorrect order of clauses

3.

What is the output of this code?

nums = {'x': 10, 'y': 5, 'z': 0}
filtered = {k: v for k, v in nums.items() if v}
print(filtered)

medium

A. {'x': 10, 'y': 5}

B. {}

C. {'z': 0}

D. {'x': 10, 'y': 5, 'z': 0}

Solution

1. Step 1: Understand the condition if v

   This condition filters out values that are 'falsy' in Python, like 0, None, or empty.

2. Step 2: Apply the condition to each item

   Values 10 and 5 are truthy, so included; 0 is falsy, so excluded.

3. Final Answer:

   {'x': 10, 'y': 5} -> Option A

4. Quick Check:

   Falsy values excluded = {'x': 10, 'y': 5} [OK]

Hint: Falsy values like 0 are skipped when condition is just 'if v' [OK]

Common Mistakes:

• Including zero as truthy
• Confusing keys and values
• Expecting all items to be included

4.

Find the error in this dictionary comprehension:

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

medium

A. No error, outputs {'b': 2, 'c': 3}

B. SyntaxError due to incorrect use of else in comprehension

C. Outputs {'a': 0, 'b': 2, 'c': 3}

D. KeyError because of else clause

Solution

1. Step 1: Check the use of else in comprehension condition

   Dictionary comprehensions cannot have an else clause directly after the if condition.

2. Step 2: Identify correct syntax

   To use else, it must be inside a value expression with a ternary operator, not after if.

3. Final Answer:

   SyntaxError due to incorrect use of else in comprehension -> Option B

4. Quick Check:

   else must be inside value expression, not after if [OK]

Hint: Use ternary inside value, not else after if in comprehension [OK]

Common Mistakes:

• Placing else after if in comprehension
• Confusing ternary operator syntax
• Expecting else to filter keys

5.

You have a dictionary of student scores:
scores = {'Alice': 85, 'Bob': 42, 'Charlie': 73, 'David': 58}
Use dictionary comprehension with a condition to create a new dictionary passed containing only students who scored 60 or more, but store their scores as 'Pass' instead of the number.
Which code correctly does this?

hard

A. {k: 'Pass' for k, v in scores.items() if v > 60}

B. {k: 'Pass' if v >= 60 else 'Fail' for k, v in scores.items()}

C. {k: v for k, v in scores.items() if v >= 60}

D. {k: 'Pass' for k, v in scores.items() if v >= 60}

Solution

1. Step 1: Understand the requirement

   We want only students with scores 60 or more, and their values replaced by 'Pass'.

2. Step 2: Check each option

   {k: 'Pass' for k, v in scores.items() if v >= 60} filters with if v >= 60 and sets value to 'Pass'. {k: 'Pass' if v >= 60 else 'Fail' for k, v in scores.items()} includes all students with ternary but no filtering. {k: v for k, v in scores.items() if v >= 60} keeps original scores. {k: 'Pass' for k, v in scores.items() if v > 60} excludes 60 exactly.

3. Final Answer:

   {k: 'Pass' for k, v in scores.items() if v >= 60} -> Option D

4. Quick Check:

   Filter and replace value with 'Pass' = {k: 'Pass' for k, v in scores.items() if v >= 60} [OK]

Hint: Filter with if, set value directly to 'Pass' in comprehension [OK]

Common Mistakes:

• Using else without filtering
• Not including scores equal to 60
• Keeping original scores instead of 'Pass'