Dictionary iteration in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of iterating dictionary keys

What is the output of this code?

my_dict = {'a': 1, 'b': 2, 'c': 3}
result = []
for key in my_dict:
    result.append(key)
print(result)

Python

my_dict = {'a': 1, 'b': 2, 'c': 3}
result = []
for key in my_dict:
    result.append(key)
print(result)

A ['a', 1, 'b', 2, 'c', 3]

B [1, 2, 3]

C ['a', 'b', 'c']

D []

💡 Hint

Iterating a dictionary by default goes over its keys.

When you loop over a dictionary, you get its keys one by one. So appending keys to a list results in ['a', 'b', 'c'].

❓ Predict Output

intermediate

Output of iterating dictionary items

What is the output of this code?

my_dict = {'x': 10, 'y': 20}
result = []
for k, v in my_dict.items():
    result.append((k, v))
print(result)

Python

my_dict = {'x': 10, 'y': 20}
result = []
for k, v in my_dict.items():
    result.append((k, v))
print(result)

A [('x', 10), ('y', 20)]

B ['x', 'y']

C [10, 20]

D [('x', 'y'), (10, 20)]

💡 Hint

items() returns key-value pairs as tuples.

The items() method returns pairs of (key, value). Appending these pairs creates a list of tuples.

❓ Predict Output

advanced

Output of modifying dictionary during iteration

What happens when you run this code?

my_dict = {'a': 1, 'b': 2}
for key in my_dict:
    my_dict['c'] = 3
print(my_dict)

Python

my_dict = {'a': 1, 'b': 2}
for key in my_dict:
    my_dict['c'] = 3
print(my_dict)

A {'a': 1, 'b': 2, 'c': 3}

B RuntimeError: dictionary changed size during iteration

C {'a': 1, 'b': 2}

D SyntaxError

💡 Hint

Changing a dictionary while looping over it causes an error.

Python does not allow changing the size of a dictionary during iteration. Adding a new key causes a RuntimeError.

🧠 Conceptual

advanced

Number of items after filtering dictionary keys

Given this code, how many items will the resulting dictionary have?

original = {'a': 1, 'b': 2, 'c': 3, 'd': 4}
filtered = {k: v for k, v in original.items() if v % 2 == 0}
print(len(filtered))

Python

original = {'a': 1, 'b': 2, 'c': 3, 'd': 4}
filtered = {k: v for k, v in original.items() if v % 2 == 0}
print(len(filtered))

A 1

B 3

C 4

D 2

💡 Hint

Count how many values are even numbers.

Only values 2 and 4 are even, so filtered dictionary has 2 items.

🔧 Debug

expert

Identify the error in dictionary iteration code

What error does this code raise?

data = {'x': 5, 'y': 10}
for key, value in data:
    print(key, value)

Python

data = {'x': 5, 'y': 10}
for key, value in data:
    print(key, value)

A ValueError: not enough values to unpack (expected 2, got 1)

B KeyError

C SyntaxError

D No error, prints keys and values

💡 Hint

Iterating a dictionary yields keys, which are strings here.

Loop tries to unpack each key (a string) into two variables, causing ValueError: not enough values to unpack (expected 2, got 1).

Practice

1. What does the following code do?
for key in my_dict:

easy

A. Iterates over the keys of the dictionary

B. Iterates over the values of the dictionary

C. Iterates over the key-value pairs as tuples

D. Raises a syntax error

Solution

1. Step 1: Understand the for loop syntax

   The loop uses for key in my_dict, which by default iterates over dictionary keys.

2. Step 2: Identify what is accessed in each iteration

   Each iteration gives one key from the dictionary, not values or pairs.

3. Final Answer:

   Iterates over the keys of the dictionary -> Option A

4. Quick Check:

   for key in dict = keys [OK]

Hint: for key in dict loops over keys only [OK]

Common Mistakes:

• Thinking it loops over values
• Confusing keys with key-value pairs
• Assuming it raises an error

2. Which of the following is the correct syntax to iterate over keys and values in a dictionary my_dict?

easy

A. for key, value in my_dict.keys():

B. for key, value in my_dict:

C. for key in my_dict.values():

D. for key, value in my_dict.items():

Solution

1. Step 1: Recall dictionary methods for iteration

   To get keys and values, use my_dict.items() which returns key-value pairs.

2. Step 2: Check syntax correctness

   Only for key, value in my_dict.items(): correctly unpacks keys and values.

3. Final Answer:

   for key, value in my_dict.items(): -> Option D

4. Quick Check:

   Use dict.items() for keys and values [OK]

Hint: Use dict.items() to get keys and values together [OK]

Common Mistakes:

• Using my_dict without .items() for key-value pairs
• Trying to unpack keys() or values() which return single elements
• Syntax errors from missing .items()

3. What is the output of this code?

my_dict = {'a': 1, 'b': 2}
for k, v in my_dict.items():
    print(k, v)

medium

A. a 1\nb 2

B. ('a', 1)\n('b', 2)

C. a\nb

D. 1\n2

Solution

1. Step 1: Understand the loop over items()

   The loop unpacks each key-value pair from my_dict.items().

2. Step 2: Analyze the print statement

   It prints the key and value separated by space on each line, so output lines are "a 1" and "b 2".

3. Final Answer:

   a 1\nb 2 -> Option A

4. Quick Check:

   print(k, v) prints key and value separated by space [OK]

Hint: print(k, v) shows key and value separated by space [OK]

Common Mistakes:

• Expecting tuples printed instead of separate values
• Printing only keys or only values
• Confusing output format with list or tuple

4. Find the error in this code:

my_dict = {'x': 10, 'y': 20}
for key, value in my_dict:
    print(key, value)

medium

A. Dictionary keys cannot be unpacked

B. Incorrect print statement syntax

C. Missing .items() after my_dict

D. No error, code runs fine

Solution

1. Step 1: Identify the iteration target

   The loop tries to unpack two variables from my_dict directly, which yields only keys.

2. Step 2: Correct the iteration method

   To unpack key and value, use my_dict.items() instead of just my_dict.

3. Final Answer:

   Missing .items() after my_dict -> Option C

4. Quick Check:

   Use my_dict.items() to unpack key, value [OK]

Hint: Use .items() to unpack key and value in loop [OK]

Common Mistakes:

• Trying to unpack keys only without .items()
• Assuming dict directly yields key-value pairs
• Ignoring error message about unpacking

5. Given the dictionary data = {'a': 0, 'b': 2, 'c': 0, 'd': 4}, which code snippet creates a new dictionary with only keys having non-zero values?

hard

A. new_dict = {k: v for k, v in data if v != 0}

B. new_dict = {k: v for k, v in data.items() if v != 0}

C. new_dict = {k: v for k in data.keys() if data[k] != 0}

D. new_dict = {k: v for v, k in data.items() if v != 0}

Solution

1. Step 1: Understand dictionary comprehension with condition

   We want to keep only items where value is not zero, so use if v != 0 in comprehension.

2. Step 2: Check correct unpacking and syntax

   new_dict = {k: v for k, v in data.items() if v != 0} correctly unpacks key and value from data.items() and filters by value.

3. Final Answer:

   new_dict = {k: v for k, v in data.items() if v != 0} -> Option B

4. Quick Check:

   Use dict.items() and filter with if condition [OK]

Hint: Use dict.items() with if condition in comprehension [OK]

Common Mistakes:

• Forgetting .items() when unpacking key and value
• Swapping key and value in unpacking
• Using keys() without accessing values