Due at 11:59pm on Tuesday, 8/1/2017.

Starter Files

Download lab12.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the OK autograder.


By the end of this lab, you should have submitted the lab with python3 ok --submit. 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 1, 2, 3, 4, 5, 6, and 7 in lab12.py and submit through OK.
  • Questions 8, 9, 10, and 11 are extra practice. They can be found in the lab12_extra.py file. It is recommended that you complete these problems on your on time.

Iterables and Iterators

In lecture, we studied several Python object interfaces, or protocols. In this lab, we will study a new protocol, the iterator protocol. Implementing this protocol allows us to use our objects in for loops! Remember the for loop? (We really hope so!)

for elem in something_iterable:
    # do something

for loops work on any object that is iterable. We previously described it as working with any sequence -- all sequences are iterable, but there are other objects that are also iterable! As it turns out, for loops are actually translated by the interpreter into the following code:

the_iterator = iter(something_iterable)
    while True:
        elem = next(the_iterator)
        # do something
except StopIteration:

That is, it first calls the built-in iter function to create an iterator, saving it in some new, hidden variable (we've called it the_iterator here). It then repeatedly calls the built-in next function on this iterator to get values of elem and stops when that function raises StopIteration.

Question 1: WWPD: Iterators

What would Python display? Try to figure it out before you type it into the interpreter!

Use OK to test your knowledge with the following "What Would Python Display?" questions:

python3 ok -q iterators -u
>>> s = [1, [2, [3, [4]]]]
>>> t = iter(s)
>>> next(t)
>>> next(iter(next(t)))
>>> list(t)
>>> next(map(lambda x: x * x, filter(lambda y: y > 4, range(10))))
>>> tuple(map(abs, reversed(range(-6, -4))))
(5, 6)
>>> r = reversed(range(10000)) >>> next(r) - next(r)
>>> xs = [2, 3, 4, 5] >>> y, z = iter(xs), iter(xs) >>> next(zip(y, z))
(2, 2)
>>> next(zip(y, y))
(3, 4)


A generator function returns a special type of iterator called a generator object. Generator functions have yield statements within the body of the function. Calling a generator function 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.
  • __next__ 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.

When a generator runs to the end of the function, it raises StopIteration.

Another useful tool for generators is the yield from statement (introduced in Python 3.3). yield from will yield all values from an iterator or iterable.

Question 2: WWPD: Generators

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

python3 ok -q generators -u
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__()
>>> 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

Question 3: Countdown

Write a generator function that counts down to 0.

def countdown(n):
    A generator that counts down from N to 0.
    >>> for number in countdown(5):
    ...     print(number)
    >>> for number in countdown(2):
    ...     print(number)
"*** YOUR CODE HERE ***"
while n >= 0: yield n n = n - 1

Use OK to test your code:

python3 ok -q countdown

Question 4: Trap

Write a generator function that yields the first k values in iterable s, but raises a ValueError exception if any more values are requested.

def trap(s, k):
    """Return a generator that yields the first K values in iterable S,
    but raises a ValueError exception if any more values are requested.

    >>> t = trap([3, 2, 1], 2)
    >>> next(t)
    >>> next(t)
    >>> next(t)
    >>> list(trap(range(5), 5))
    assert len(s) >= k
"*** YOUR CODE HERE ***"
t = iter(s) while k > 0: yield next(t) k -= 1 raise ValueError("It's a trap!")

Use OK to test your code:

python3 ok -q trap

Question 5: Repeated

Implement a function (not a generator function) that returns the first value in iterable t that appears k times in a row.

def repeated(t, k):
    """Return the first value in iterable T that appears K times in a row.

    >>> s = [3, 2, 1, 2, 1, 4, 4, 5, 5, 5]
    >>> repeated(trap(s, 7), 2)
    >>> repeated(trap(s, 10), 3)
    >>> print(repeated([4, None, None, None], 3))
    assert k > 1
"*** YOUR CODE HERE ***"
count = 0 last_item = None for item in t: if item == last_item: count += 1 else: last_item = item count = 1 if count == k: return item

Use OK to test your code:

python3 ok -q repeated


The Stream class defines a lazy sequence, a lazily evaluated linked list. In other words, a Stream's elements (except for the first element) are only evaluated as those values are needed.

class Stream:
    empty = 'empty'
    def __init__(self, first, compute_rest=lambda: Stream.empty):
        self.first = first
        self.cached_rest = None
        assert callable(compute_rest)
        self.compute_rest = compute_rest
    def rest(self):
        """Return the rest, computing it if necessary."""
        if self.compute_rest is not None:
            self.cached_rest = self.compute_rest()
            self.compute_rest = None
        return self.cached_rest
    def __repr__(self):
        rest = self.cached_rest if self.compute_rest is None else '<...>'
        return 'Stream({}, {})'.format(self.first, rest)

Instead of specifying all of the elements in __init__, we provide a function, compute_rest, that encapsulates the algorithm used to calculate the remaining elements of the stream. The code in the function body is not evaluated until it is called, which lets us implement the desired evaluation behavior.

This implementation of streams also uses memoization. The first time a program asks a Stream for its rest field, the Stream code computes the required value using compute_rest, saves the resulting value, and then returns it. After that, every time the rest field is referenced, the stored value is simply returned and it is not computed again.

Here is an example (which you may use in solutions):

def make_integer_stream(first=1):
    def compute_rest():
        return make_integer_stream(first+1)
    return Stream(first, compute_rest)

We start out with a stream whose first element is 1, and whose compute_rest function creates another stream. So when we do compute the rest, we get another stream whose first element is one greater than the previous element, and whose compute_rest creates another stream. Hence, we effectively get an infinite stream of integers, computed one at a time. This is almost like an infinite recursion, but one which can be viewed one step at a time, and so does not crash.

Another example:

def map_stream(fn, s):
    if s is Stream.empty:
        return s
    return Stream(fn(s.first), lambda: map_stream(fn, s.rest))

Question 6: Ones

Define a stream ones that creates an infinite stream of ones.

ones = None
ones = Stream(1, lambda: ones)
def ones_test(): """ >>> ones.first, ones.rest.first, ones.rest.rest.first, ones.rest.rest.rest.first (1, 1, 1, 1) """

Use OK to test your code:

python3 ok -q ones_test

Question 7: Scan

Define a function scan, which takes in a two argument function f, an initial value a0, and an infinite stream with elements

a1, a2, a3, ...

and outputs the stream

a0, f(a0, a1), f(f(a0, a1), a2), f(f(f(a0, a1), a2), a3), ...
def scan(f, initial_value, stream):
    >>> ones = Stream(1, lambda: ones)
    >>> naturals = scan(lambda x, y: x + y, 1, ones)
    >>> _ = naturals.rest.rest.rest
    >>> naturals
    Stream(1, Stream(2, Stream(3, Stream(4, <...>))))
    >>> factorials = scan(lambda x, y: x * y, 1, naturals)
    >>> _ = factorials.rest.rest.rest.rest
    >>> factorials
    Stream(1, Stream(1, Stream(2, Stream(6, Stream(24, <...>)))))
"*** YOUR CODE HERE ***"
return Stream(initial_value, lambda: scan(f, f(initial_value, stream.first), stream.rest))

Use OK to test your code:

python3 ok -q scan

Extra Questions

Question 8: Scale

Implement the generator function scale(s, k), which yields elements of the given iterable s, scaled by k.

def scale(s, k):
    """Yield elements of the iterable s scaled by a number k.

    >>> s = scale([1, 5, 2], 5)
    >>> type(s)
    <class 'generator'>
    >>> list(s)
    [5, 25, 10]

    >>> m = scale(naturals(), 2)
    >>> [next(m) for _ in range(5)]
    [2, 4, 6, 8, 10]
"*** YOUR CODE HERE ***"
for elem in s: yield elem * k

Use OK to test your code:

python3 ok -q scale

Question 9: Remainder Generator

Like functions, generators can also be higher-order. For this problem, we will be writing remainders_generator, which yields a series of generator objects.

remainders_generator takes in an integer m, and yields m different generators. The first generator is a generator of multiples of m, i.e. numbers where the remainder is 0. The second, a generator of natural numbers with remainder 1 when divided by m. The last generator yield natural numbers with remainder m - 1 when divided by m.

def remainders_generator(m):
    Takes in an integer m, and yields m different remainder groups
    of m.

    >>> remainders_mod_four = remainders_generator(4)
    >>> for rem_group in remainders_mod_four:
    ...     for _ in range(3):
    ...         print(next(rem_group))
"*** YOUR CODE HERE ***"
def remainder_group(rem): start = rem while True: yield start start += m for rem in range(m): yield remainder_group(rem)

Note that if you have implemented this correctly, each of the generators yielded by remainder_generator will be infinite - you can keep calling next on them forever without running into a StopIteration exception.

Hint: Consider defining an inner generator function. What arguments should it take in? Where should you call it?

Use OK to test your code:

python3 ok -q remainders_generator

Question 10: Zip generator

For this problem, we will be writing zip_generator, which yields a series of lists, each containing the nth items of each iterable. It should stop when the smallest iterable runs out of elements.

def zip(*iterables):
    Takes in any number of iterables and zips them together.
    Returns a generator that outputs a series of lists, each
    containing the nth items of each iterable.
    >>> z = zip([1, 2, 3], [4, 5, 6], [7, 8])
    >>> for i in z:
    ...     print(i)
    [1, 4, 7]
    [2, 5, 8]
"*** YOUR CODE HERE ***"
iterators = [iter(iterable) for iterable in iterables] while True: yield [next(iterator) for iterator in iterators]

Use OK to test your code:

python3 ok -q zip

Question 11: Convolve Streams

Define a function convolve_streams, which takes in two streams a and b, each of which represents a polynomial (the nth term represents the coefficient of x^n in the polynomial). For example, if we have the streams

2, 3, 1, 0, 0, 0, 0...


6, -1, 0, 0, 0, 0...

convolve_streams should output the stream

12, 16, 3, -1, 0, 0, 0...

Hint: note that (a0 + a1 * x + a2 * x^2...)b(x) = a0 b(x) + (a1 * x + a2 * x^2...)b(x)

def convolve_streams(a, b):
    >>> zeros = Stream(0, lambda: zeros)
    >>> a = Stream(2, lambda: Stream(3, lambda: Stream(1, lambda: zeros)))
    >>> b = Stream(6, lambda: Stream(-1, lambda: zeros))
    >>> c = convolve_streams(a, b)
    >>> _ = c.rest.rest.rest.rest
    >>> c
    Stream(12, Stream(16, Stream(3, Stream(-1, Stream(0, <...>)))))
"*** YOUR CODE HERE ***"
def add_streams(u, v): return Stream(u.first + v.first, lambda: add_streams(u.rest, v.rest)) def scale_stream(k, u): return Stream(k * u.first, lambda: scale_stream(k, u.rest)) return add_streams(scale_stream(a.first, b), Stream(0, lambda: convolve_streams(a.rest, b)))

Use OK to test your code:

python3 ok -q convolve_streams