Basic dictionary comprehension 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 '+' instead of '**' will add 2 to x, not square it.

Using '*' will multiply x by 2, not square it.

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 with words as keys and their lengths as values.

Python

words = ['apple', 'banana', 'cherry']
lengths = {word: [1] for word in words}
print(lengths)

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

Alen(word)

Bword

Cword.upper()

Dword[0]

Attempts:

3 left

💡 Hint

Common Mistakes

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

Using word.upper() returns the uppercase word, not its length.

The len() function returns the length of a string. Here, it gives the length of each word.

3fill in blank

hard

Fix the error in the dictionary comprehension to create a dictionary with numbers as keys and their cubes as values.

Python

cubes = {n: n[1]3 for n in range(1, 5)}
print(cubes)

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

A*

B**

C+

D//

Attempts:

3 left

💡 Hint

Common Mistakes

Using '*' multiplies n by 3 instead of cubing it.

Using '+' adds 3 to n instead of cubing it.

The exponentiation operator ** is needed to cube the number (raise it to the power of 3).

4fill in blank

hard

Complete the code to create a dictionary with words as keys and their first letters as values, but only for words longer than 5 letters.

Python

words = 'apple', 'banana', 'cherry', 'date']
result = {word: word[0 for word in words if len(word) [1] 5}
print(result)

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

A[

B>

C<

D]

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of brackets for indexing.

Using '<' instead of '>' in the condition.

The condition len(word) > 5 filters words longer than 5 letters.

5fill in blank

hard

Fill both blanks to create a dictionary with uppercase words as keys and their lengths as values, but only for words with length less than 6.

Python

words = ['apple', 'banana', 'cherry', 'date']
result = {word.upper(): {BLANK_2}} for word in words if len(word) {{BLANK_2}} 6
print(result)

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

A{word.upper()}

Blen(word)

C<

D{

Attempts:

3 left

💡 Hint

Common Mistakes

Forgetting the opening brace for the dictionary comprehension.

Using '>' instead of '<' in the condition.

Using the word itself instead of its length for values.

The dictionary comprehension starts with {. The keys are uppercase words using word.upper() (inside braces), the values are lengths with len(word), and the condition filters words with length less than 6 using <.

Practice

1. What does the following dictionary comprehension do?
{x: x*x for x in range(3)}

easy

A. Creates a list of squares from 0 to 2

B. Creates a set of squares from 0 to 2

C. Creates a dictionary with numbers 0 to 2 as keys and their squares as values

D. Creates a dictionary with keys as squares and values as numbers

Solution

1. Step 1: Understand the loop and key-value pairs

   The comprehension loops over numbers 0, 1, 2 and uses each number as the key.

2. Step 2: Calculate the value for each key

   Each value is the square of the key (x*x), so values are 0, 1, 4 respectively.

3. Final Answer:

   Creates a dictionary with numbers 0 to 2 as keys and their squares as values -> Option C

4. Quick Check:

   Dictionary comprehension = key: x, value: x*x [OK]

Hint: Look for key:value pairs inside braces with a for loop [OK]

Common Mistakes:

• Confusing dictionary with list or set comprehension
• Mixing keys and values in comprehension
• Thinking it creates a list instead of a dictionary

2. Which of the following is the correct syntax for a dictionary comprehension that creates keys from 1 to 3 and values as their double?

easy

A. [x: 2*x for x in range(1, 4)]

B. {x: 2*x for x in range(1, 4)}

C. {x => 2*x for x in range(1, 4)}

D. (x: 2*x for x in range(1, 4))

Solution

1. Step 1: Identify correct dictionary comprehension syntax

   Dictionary comprehensions use curly braces with key:value pairs and a for loop inside.

2. Step 2: Check each option

   {x: 2*x for x in range(1, 4)} uses curly braces and colon correctly. Options B, C, D use wrong brackets or arrow syntax.

3. Final Answer:

   {x: 2*x for x in range(1, 4)} -> Option B

4. Quick Check:

   Curly braces + key:value + for loop = correct syntax [OK]

Hint: Remember dictionary comprehension uses curly braces and colon [OK]

Common Mistakes:

• Using square brackets instead of curly braces
• Using arrow (=>) instead of colon for key-value
• Using parentheses which create generator expressions

3. What is the output of this code?

nums = [1, 2, 3]
squares = {n: n**2 for n in nums if n % 2 != 0}
print(squares)

medium

A. {1: 1, 3: 9}

B. {2: 4}

C. {1: 1, 2: 4, 3: 9}

D. {}

Solution

1. Step 1: Understand the filtering condition

   The comprehension includes only numbers where n % 2 != 0, meaning odd numbers only (1 and 3).

2. Step 2: Calculate squares for filtered numbers

   Squares are 1*1=1 and 3*3=9, so dictionary is {1:1, 3:9}.

3. Final Answer:

   {1: 1, 3: 9} -> Option A

4. Quick Check:

   Filter odd numbers and square them = {1:1, 3:9} [OK]

Hint: Check the if condition filters keys before squaring [OK]

Common Mistakes:

• Including even numbers by mistake
• Confusing list comprehension output with dictionary
• Ignoring the if condition in comprehension

4. Find the error in this dictionary comprehension:

result = {x, x*2 for x in range(3)}

medium

A. Missing colon between key and value

B. Using parentheses instead of curly braces

C. Range function is incorrect

D. No error, code is correct

Solution

1. Step 1: Check syntax for key-value pairs

   Dictionary comprehension requires a colon ':' between key and value, but here a comma ',' is used.

2. Step 2: Confirm other parts are correct

   Curly braces and range(3) are correct, so the error is the missing colon.

3. Final Answer:

   Missing colon between key and value -> Option A

4. Quick Check:

   Key:value pairs need colon, not comma [OK]

Hint: Use colon ':' to separate key and value in dict comprehension [OK]

Common Mistakes:

• Using comma instead of colon between key and value
• Confusing set comprehension with dictionary comprehension
• Using wrong brackets for comprehension

5. Given a list of words, create a dictionary comprehension that maps each word to its length but only include words longer than 3 letters.
Which code correctly does this?

words = ['cat', 'house', 'dog', 'elephant']

hard

A. {word: len(word) for word in words where len(word) > 3}

B. {word: len(word) for word in words if word > 3}

C. {word: len(word) if len(word) > 3 for word in words}

D. {word: len(word) for word in words if len(word) > 3}

Solution

1. Step 1: Understand filtering condition syntax

   The if condition must come after the for loop to filter words by length correctly.

2. Step 2: Check each option's syntax

   {word: len(word) for word in words if len(word) > 3} correctly places 'if len(word) > 3' after the for loop. Others have syntax errors or wrong condition placement.

3. Final Answer:

   {word: len(word) for word in words if len(word) > 3} -> Option D

4. Quick Check:

   Filter after for loop with if condition [OK]

Hint: Put if condition after for loop in dictionary comprehension [OK]

Common Mistakes:

• Placing if condition before for loop
• Using invalid syntax like 'where' instead of 'if'
• Comparing word string directly to number