Lab 11: Iterators and Generators

Due at 9:00pm on 11/15/2018.

Starter Files

Download lab11.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the OK autograder.

Submission

When you are done, submit the lab by uploading the lab11.py file to okpy.org. You may submit more than once before the deadline; only the final submission will be graded.

• To receive credit for this lab, you must complete Questions 2, 3, 4, and 5 in lab11.py and submit through OK.

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.

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 --local

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

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

    def __iter__(self):
        return self

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

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

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

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

    def __iter__(self):
        return self

Question 2: Restart

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

python3 ok -q restart -u --local

Try this!

>>> iterator = IteratorA()
>>> [num for num in iterator]

Then again:

>>> [num for num in iterator]

This happens because the instance variables are not reset each time a for loop is started. Therefore, when a StopIteration exception is raised at the end of the first for loop it certainly will be raised immediately at the beginning of the second. With that in mind, try writing an iterator that "restarts" every time it is run through a for loop.

class IteratorRestart:
    """
    >>> iterator = IteratorRestart(2, 7)
    >>> for num in iterator:
    ...     print(num)
    2
    3
    4
    5
    6
    7
    >>> for num in iterator:
    ...     print(num)
    2
    3
    4
    5
    6
    7
    """
    def __init__(self, start, end):
        "*** YOUR CODE HERE ***"
        self.start = start
        self.end = end
        self.current = start
    def __next__(self):
        "*** YOUR CODE HERE ***"
        if self.current > self.end:
            raise StopIteration
        self.current += 1
        return self.current - 1
    def __iter__(self):
        "*** YOUR CODE HERE ***"
        self.current = self.start
        return self

Use OK to test your code:

python3 ok -q IteratorRestart --local

Question 3: 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)

class Str:
    """
    >>> s = Str("hello")
    >>> for char in s:
    ...     print(char)
    ...
    h
    e
    l
    l
    o
    >>> for char in s:    # a standard iterator does not restart
    ...     print(char)
    """
    "*** 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]

Use OK to test your code:

python3 ok -q Str --local

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?

Question 4: 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 --local

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 --local

Question 5: 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 --local