Iterator protocol in Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Iterator protocol in Python?

The Iterator protocol is a way Python objects provide a standard way to access elements one at a time, using two methods: __iter__() and __next__().

beginner

What does the __iter__() method do?

The __iter__() method returns the iterator object itself. It allows an object to be used in a loop like for.

beginner

What happens when __next__() is called on an iterator?

The __next__() method returns the next item from the container. If there are no more items, it raises a StopIteration exception to signal the end.

intermediate

How does a for loop use the Iterator protocol internally?

A for loop calls __iter__() to get an iterator, then repeatedly calls __next__() to get each item until StopIteration is raised.

intermediate

How can you make a custom object iterable using the Iterator protocol?

Define __iter__() to return an iterator object, and define __next__() in that iterator to return items one by one, raising StopIteration when done.

Which method must an iterator implement to return the next item?

A __next__()

B __iter__()

C __getitem__()

D __call__()

The __next__() method returns the next item or raises StopIteration when done.

What does the __iter__() method return?

A The iterator object itself

B A list of all items

C The first item

D None

__iter__() returns the iterator object to be used in loops.

What exception signals the end of iteration?

A TypeError

B IndexError

C ValueError

D StopIteration

StopIteration tells Python that no more items are available.

Which of these is true about an iterable?

A It raises StopIteration immediately

B It has an __iter__() method

C It has a __next__() method

D It is always a list

An iterable must have an __iter__() method that returns an iterator.

What happens if you call next() on an iterator with no more items?

A Raises TypeError

B Returns None

C Raises StopIteration

D Returns the last item again

Calling next() on an exhausted iterator raises StopIteration.

Explain how the Iterator protocol works in Python and why it is useful.

Think about how Python loops over things like lists or files.

Describe how you would create a custom iterator for your own object.

Focus on the two special methods needed.

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