Dictionary comprehension with condition in Python - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a dictionary with numbers as keys and their squares as values.

Python

squares = {x: x[1]2 for x in range(1, 6)}
print(squares)

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

A**

B+

C*

D//

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication (*) instead of exponentiation (**).

Using addition (+) which adds numbers instead of squaring.

The ** operator is used to raise a number to a power. Here, x**2 means x squared.

2fill in blank

medium

Complete the code to create a dictionary of numbers and their squares only for numbers greater than 3.

Python

filtered_squares = {x: x**2 for x in range(1, 7) if x [1] 3}
print(filtered_squares)

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

A==

B!=

C>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using less than (<) which selects numbers smaller than 3.

Using equality (==) which selects only number 3.

The condition x > 3 filters numbers greater than 3.

3fill in blank

hard

Fix the error in the dictionary comprehension to include only even numbers and their squares.

Python

even_squares = {n: n**2 for n in range(1, 10) if n [1] 2 == 0}
print(even_squares)

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

A//

B%

C*

D**

Attempts:

3 left

💡 Hint

Common Mistakes

Using floor division (//) which does not check for evenness.

Using multiplication (*) or exponentiation (**) which are unrelated.

The modulo operator % gives the remainder. n % 2 == 0 checks if n is even.

4fill in blank

hard

Fill both blanks to create a dictionary of words and their lengths only if the length is greater than 4.

Python

word_lengths = {word: [1] for word in words if len(word) [2] 4}

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

Alen(word)

Bword

C>

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using the word itself as value instead of its length.

Using less than or equal (<=) which selects shorter words.

The value should be the length of the word using len(word). The condition should check if length is greater than 4 using >.

5fill in blank

hard

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

Python

result = { [1]: [2] for word in words if len(word) [3] 3 }

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

Aword.upper()

Blen(word)

C>

Dword

Attempts:

3 left

💡 Hint

Common Mistakes

Using the original word as key instead of uppercase.

Using less than or equal (<=) which includes shorter words.

The key should be the uppercase word using word.upper(). The value is the length len(word). The condition filters words longer than 3 using >.

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'