Iterables and Iterators

Remember the for loop? (We really hope so.) The object the for loop iterates over must be an iterable! An iterable is a way of representing explicit sequences (like lists or strings) as well as implicit sequences (like the natural numbers 1, 2, 3, ...).

for elem in iterable:
    # do something

for loops only work with iterables. This means the object you want to use a for loop on must implement the iterable interface. To implement the iterable interface, an object must define an __iter__ method that returns an object that implements the iterator interface. To implement the iterator interface, an object must define a __next__ method to compute and return the next element in the sequence. If the iterator exhausts the sequence, it raises StopIteration to send a signal to indicate that it reaches the end.

An iterable object can create an arbitrary amount of iterator objects. In addition, the iterators are independent of each other; in other words they can have a different position in the sequence.

Here is a table summarizing the required methods of the iterable and iterator interfaces/protocols. Python also has more documentation about iterator types.

Iterable Iterator
__iter__: return a new iterator __iter__: must return itself
__next__: return the next element, or raise StopIteration

In Python, an iterator must also be an iterable. In other words, it must have a __iter__ method that returns itself (with the current state unaltered).

Analogy: an iterable is like a book (one can flip through the pages) and an iterator is a bookmark (saves the position and can locate the next page). Calling __iter__ on a book gives you a new bookmark independent of other bookmarks, but calling __iter__ on a bookmark gives you the bookmark itself, without changing its position at all.

Here is an example of a class definition for an object that implements the iterator interface:

class AnIterator:
    def __init__(self):
        self.current = 0

    def __next__(self):
        if self.current > 5:
            raise StopIteration
        self.current += 1
        return self.current

    def __iter__(self):
        return self

Let's go ahead and try out our new toy.

>>> for num in AnIterator():
...     print(num)
1
2
3
4
5
6

This is somewhat equivalent to running:

t = AnIterator()
t = iter(t) # iter(t) is the same as t.__iter__()
try:
    while True:
        # next(t) is the same as t.__next__()
        print(next(t))
except StopIteration as e:
    pass

Question 1: Does it work?

Consider the following iterators. Which ones work and which ones don't? Why?

Use OK to test your knowledge with the following conceptual questions:

python3 ok -q does_it_work -u
class IteratorA:
    def __init__(self):
        self.start = 10

    def __next__(self):
        if self.start > 100:
            raise StopIteration
        self.start += 20
        return self.start

    def __iter__(self):
        return self

No problem, this is a beautiful iterator.

class IteratorB:
    def __init__(self):
        self.start = 5

    def __iter__(self):
        return self

Oh no! Where is __next__? This fails to implement the iterator interface because calling __iter__ doesn't return something that has a __next__ method.

class IteratorC:
    def __init__(self):
        self.start = 5

    def __next__(self):
        if self.start == 10:
            raise StopIteration
        self.start += 1
        return self.start

This also fails to implement the iterator interface. Without the __iter__ method, the for loop will error. The for loop needs to call __iter__ first because some objects might not implement the __next__ method themselves, but calling __iter__ will return an object that does.

class IteratorD:
    def __init__(self):
        self.start = 1

    def __next__(self):
        self.start += 1
        return self.start

    def __iter__(self):
        return self

This is technically an iterator, because it implements both __iter__ and __next__. Notice that it's an infinite sequence! Sequences like these are the reason iterators are useful. Because iterators delay computation, we can use a finite amount of memory to represent an infinitely long sequence.

Question 2: Str

Write an iterator that takes a string as input and outputs the letters in order when iterated over.

class Str:
    def __init__(self, str):
        self.str = str
    def __iter__(self):
        return iter(self.str)

That works (why?), but just kidding.

class Str:
"*** YOUR CODE HERE ***"
def __init__(self, str): self.str = str self.i = -1 def __iter__(self): return self def __next__(self): self.i += 1 if self.i >= len(self.str): raise StopIteration return self.str[self.i]

Test your code by either writing your own doc tests, or by running python3 -i lab11.py

There is no doctest for this problem. Test this on your own! We recommend starting with an interactive python shell python3 -i lab11.py.

Generators

A generator function returns a special type of iterator called a generator object. Such functions can be written using a yield statement:

def <generator_fn_name>():
    <somevariable> = <something>
    while <predicate>:
        yield <something>
        <increment somevariable>

Calling a generator function (a function with a yield statement in it) makes it return a generator object rather than executing the body of the function.

The reason we say a generator object is a special type of iterator is that it has all the properties of an iterator, meaning that:

  • Calling the __iter__ method makes a generator object return itself without modifying its current state.
  • Calling the __next__ method makes a generator object compute and return the next object in its sequence. If the sequence is exhausted, StopIteration is raised.
  • Typically, a generator should not restart unless it's defined that way. But calling the generator function returns a brand new generator object (like calling __iter__ on an iterable object).

However, they do have some fundamental differences:

  • An iterator is a class with __next__ and __iter__ explicitly defined, but a generator can be written as a mere function with a yield in it.
  • __iter__ in an iterator uses return, but a generator uses yield.
  • A generator "remembers" its state for the next __next__ call. Therefore, the first __next__ call works like this:

    1. Enter the function, run until the line with yield.
    2. Return the value in the yield statement, but remember the state of the function for future __next__ calls.

    And subsequent __next__ calls work like this:

    1. Re-enter the function, start at the line after yield, and run until the next yield statement.
    2. Return the value in the yield statement, but remember the state of the function for future __next__ calls.

Use OK to test your knowledge with the following What would Python print questions:

python3 ok -q generators -u --local
def generator():
    print("Starting here")
    i = 0
    while i < 6:
        print("Before yield")
        yield i
        print("After yield")
        i += 1
>>> g = generator()
>>> g # what type of object is this?
______
<generator object ...>
>>> g == iter(g) # equivalent of g.__iter__()
______
True
>>> next(g) # equivalent of g.__next__()
______
Starting here Before yield 0
>>> next(g)
______
After yield Before yield 1
>>> next(g)
______
After yield Before yield 2

Trace through the code and make sure you know where and why each statement is printed.

You might have noticed from the Iterators section that IteratorB, which didn't define a __next__ method, failed to run in the for loop. However, this is not always the case.

class IterGen:
    def __init__(self):
        self.start = 5

    def __iter__(self):
        while self.start < 10:
            self.start += 1
            yield self.start

for i in IterGen():
    print(i)

Why does this iterable work without defining a __next__ method?

The for loop only expects the object returned by __iter__ to have a __next__ method. The __iter__ method is a generator function because of the yield statement in the body. Therefore, when __iter__ is called, it returns a generator object, which you can call __next__ on.

Question 3: Countdown

Write a generator that counts down to 0.

Write it in both ways: using a generator function on its own, and within the __iter__ method of a class.

def countdown(n):
    """
    >>> from types import GeneratorType
    >>> type(countdown(0)) is GeneratorType # countdown is a generator
    True
    >>> for number in countdown(5):
    ...     print(number)
    ...
    5
    4
    3
    2
    1
    0
    """
"*** YOUR CODE HERE ***"
while n >= 0: yield n n = n - 1

Use OK to test your code:

python3 ok -q countdown
class Countdown:
    """
    >>> from types import GeneratorType
    >>> type(Countdown(0)) is GeneratorType # Countdown is an iterator
    False
    >>> for number in Countdown(5):
    ...     print(number)
    ...
    5
    4
    3
    2
    1
    0
    """
"*** YOUR CODE HERE ***"
def __init__(self, cur): self.cur = cur def __iter__(self): return self def __next__(self): result = self.cur if result < 0: raise StopIteration self.cur -= 1 return result # Alternate solution that makes __iter__ a generator function: class Countdown: def __init__(self, cur): self.cur = cur def __iter__(self): while self.cur >= 0: yield self.cur self.cur -= 1

Hint: Is it possible to not have a __next__ method in Countdown?

Use OK to test your code:

python3 ok -q Countdown

Question 4: Hailstone

Write a generator that outputs the hailstone sequence from Lab 01.

Here's a quick remainder of how the hailstone sequence is defined:

  1. Pick a positive integer n as the start.
  2. If n is even, divide it by 2.
  3. If n is odd, multiply it by 3 and add 1.
  4. Continue this process until n is 1.
def hailstone(n):
    """
    >>> for num in hailstone(10):
    ...     print(num)
    ...
    10
    5
    16
    8
    4
    2
    1
    """
"*** YOUR CODE HERE ***"
i = n while i > 1: yield i if i % 2 == 0: i //= 2 else: i = i * 3 + 1 yield i

Use OK to test your code:

python3 ok -q hailstone

Submit

Make sure to submit this assignment by running:

python3 ok --submit

Extra Credit Practice Open in a new window

These questions are new this semester. They're a mix of Parsons Problems, Code Tracing questions, and Code Writing questions.

Confused about how to use the tool? Check out https://codestyle.herokuapp.com/cs88-lab01 for some problems designed to demonstrate how to solve these types of problems.

These cover some similar material to lab, so can be helpful to further review or try to learn the material. Unlike lab and homework, after you've worked for long enough and tested your code enough times on any of these questions, you'll have the option to view an instructor solution. You'll unlock each question one at a time, either by correctly answering the previous question or by viewing an instructor solution.

Starting from lab 2 onward, each set of questions are worth half (0.5) a point per lab, for a total opportunity of 4-5 points throughout the semester.

Use OK to test your code:

python3 ok -q extra_credit