Iterator protocol in Python - Mini Project: Build & Apply

Iterator Protocol

📖 Scenario: You are working on a simple program that processes a list of fruits one by one. To do this, you will use Python's iterator protocol, which lets you manually get each item from a list.

🎯 Goal: Build a program that creates a list of fruits, gets an iterator for that list, manually retrieves each fruit using the iterator, and prints each fruit.

📋 What You'll Learn

Create a list called fruits with the exact items: 'apple', 'banana', 'cherry'

Create a variable called fruit_iterator that gets the iterator of the fruits list using the iter() function

Use the next() function to get each fruit from fruit_iterator and store them in variables first_fruit, second_fruit, and third_fruit

Print each fruit variable on its own line in the order they were retrieved

💡 Why This Matters

🌍 Real World

Iterators let programs handle data one piece at a time, which is useful when working with large files, streams, or user input step-by-step.

💼 Career

Understanding the iterator protocol is important for Python developers because many built-in functions and libraries use iterators to process data efficiently.

1

Create the list of fruits

Create a list called fruits with these exact items in order: 'apple', 'banana', 'cherry'

Python

# Create the list fruits with 'apple', 'banana', 'cherry'
# Your code here

Hint

Use square brackets [] to create a list and separate items with commas.

2

Get an iterator for the fruits list

Create a variable called fruit_iterator and set it to the iterator of the fruits list using the iter() function

Python

fruits = ['apple', 'banana', 'cherry']
# Create fruit_iterator using iter() on fruits
# Your code here

Hint

Use iter(fruits) to get the iterator.

3

Retrieve fruits using next()

Use the next() function to get each fruit from fruit_iterator and store them in variables first_fruit, second_fruit, and third_fruit in that order

Python

fruits = ['apple', 'banana', 'cherry']
fruit_iterator = iter(fruits)
# Use next() to get first_fruit, second_fruit, third_fruit
# Your code here

Hint

Call next(fruit_iterator) three times and assign each result to the correct variable.

4

Print the fruits

Print the variables first_fruit, second_fruit, and third_fruit each on its own line in that order

Python

fruits = ['apple', 'banana', 'cherry']
fruit_iterator = iter(fruits)
first_fruit = next(fruit_iterator)
second_fruit = next(fruit_iterator)
third_fruit = next(fruit_iterator)
# Print each fruit on its own line
# Your code here

Hint

Use three print() statements, one for each fruit variable.

Practice

1. What does the __iter__ method do in the iterator protocol?

easy

A. Returns the iterator object itself

B. Returns the next item in the sequence

C. Stops the iteration

D. Creates a list from the iterable

Solution

1. Step 1: Understand the role of __iter__

   The __iter__ method is called to get an iterator object from an iterable.

2. Step 2: Identify what __iter__ returns

   It returns the iterator object itself, which has the __next__ method to fetch items.

3. Final Answer:

   Returns the iterator object itself -> Option A

4. Quick Check:

   __iter__ returns iterator object [OK]

Hint: Remember: __iter__ returns the iterator itself [OK]

Common Mistakes:

• Confusing __iter__ with __next__
• Thinking __iter__ returns the next item
• Assuming __iter__ stops iteration

2. Which of the following is the correct way to define an iterator class in Python?

easy

A. class MyIter:
def __next__(self):
pass

B. class MyIter:
def next(self):
pass

C. class MyIter:
def __iter__(self):
return self
def __next__(self):
pass

D. class MyIter:
def iter(self):
return self

Solution

1. Step 1: Check required methods for iterator

   An iterator class must have __iter__ returning self and __next__ to get next item.

2. Step 2: Match methods with options

   class MyIter: def __iter__(self): return self def __next__(self): pass correctly defines both __iter__ and __next__ methods.

3. Final Answer:

   Defines both __iter__ and __next__ methods -> Option C

4. Quick Check:

   Iterator class needs __iter__ and __next__ [OK]

Hint: Iterator class must have __iter__ and __next__ methods [OK]

Common Mistakes:

• Using next() instead of __next__()
• Missing __iter__ method
• Defining iter() instead of __iter__()

3. What will be the output of this code?

class Count:
    def __init__(self, limit):
        self.limit = limit
        self.num = 0
    def __iter__(self):
        return self
    def __next__(self):
        if self.num < self.limit:
            self.num += 1
            return self.num
        else:
            raise StopIteration

for i in Count(3):
    print(i, end=' ')

medium

A. Error: StopIteration not handled

B. 0 1 2

C. 1 2 3 4

D. 1 2 3

Solution

1. Step 1: Understand the iterator behavior

   The Count class starts num at 0 and returns num+1 until it reaches limit 3.

2. Step 2: Trace the loop output

   Loop prints 1, 2, 3 then raises StopIteration to end loop.

3. Final Answer:

   1 2 3 -> Option D

4. Quick Check:

   Count(3) yields 1 to 3 [OK]

Hint: StopIteration ends loop; count from 1 to limit [OK]

Common Mistakes:

• Starting count from 0 instead of 1
• Expecting 4 as output
• Thinking StopIteration causes error

4. Identify the error in this iterator implementation:

class MyIter:
    def __init__(self):
        self.data = [1, 2, 3]
        self.index = 0
    def __iter__(self):
        return self
    def __next__(self):
        if self.index <= len(self.data):
            result = self.data[self.index]
            self.index += 1
            return result
        else:
            raise StopIteration

medium

A. The condition should be self.index < len(self.data)

B. Missing return self in __iter__

C. Should raise StopIteration before returning result

D. Index should start at 1, not 0

Solution

1. Step 1: Analyze the index condition

   Index goes from 0 to len(data)-1. Using <= allows index == len(data), causing IndexError.

2. Step 2: Correct the condition

   Change condition to self.index < len(self.data) to avoid out-of-range access.

3. Final Answer:

   The condition should be self.index < len(self.data) -> Option A

4. Quick Check:

   Index must be less than length to avoid error [OK]

Hint: Use < to avoid index out of range errors [OK]

Common Mistakes:

• Using <= instead of < in index check
• Forgetting to return self in __iter__
• Starting index at 1 instead of 0

5. You want to create a custom iterator that returns only even numbers from 0 up to a given limit (exclusive). Which implementation correctly follows the iterator protocol and filters evens?

class EvenIterator:
    def __init__(self, limit):
        self.limit = limit
        self.current = 0
    def __iter__(self):
        return self
    def __next__(self):
        while self.current < self.limit:
            val = self.current
            self.current += 1
            if val % 2 == 0:
                return val
        raise StopIteration

hard

A. Fails because __iter__ should return a new object each time

B. Correct implementation returning even numbers up to limit

C. Incorrect because it returns odd numbers instead

D. Raises StopIteration too early, missing some evens

Solution

1. Step 1: Check iterator protocol methods

   Class defines __iter__ returning self and __next__ with loop and StopIteration.

2. Step 2: Verify filtering logic

   Inside __next__, it loops until limit, returns only even values, skipping odds.

3. Final Answer:

   Correct implementation returning even numbers up to limit -> Option B

4. Quick Check:

   Iterator filters evens correctly [OK]

Hint: Use while loop inside __next__ to skip unwanted items [OK]

Common Mistakes:

• Returning odd numbers by mistake
• Not raising StopIteration when done
• Returning new iterator in __iter__ each time