Lab 10: Iterators and Generators

• lab10.zip

Due at 11:59pm on Friday, 07/27/2018.

Lab Check-in 6 questions here.

Starter Files

Download lab10.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the Ok autograder.

Submission

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. Check that you have successfully submitted your code on okpy.org.

• To receive credit for this lab, you must complete Questions 1-4 and submit through OK. Starter code for Questions 3-4 can be found in lab10.py.
• Questions 5-9 are extra practice. They can be found in the lab10_extra.py file. It is recommended that you complete these problems on your own time.

Topics

Iterables and Iterators

An iterable is any object that can be iterated through, or gone through one element at a time. One construct that we've used to iterate through an iterable is a for loop:

for elem in 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! We define an iterable as an object on which calling the built-in function iter function returns an iterator. An iterator is another type of object that allows us to iterate through an iterable by keeping track of which element is next in the sequence.

To illustrate this, consider the following block of code, which does the exact same thing as a the for statement above:

iterator = iter(iterable)
try:
    while True:
        elem = next(iterator)
        # do something
except StopIteration:
    pass

Here's a breakdown of what's happening:

• First, the built-in iter function is called on the iterable to create a corresponding iterator.
• To get the next element in the sequence, the built-in next function is called on this iterator.
• When next is called but there are no elements left in the iterator, a StopIteration error is raised. In the for loop construct, this exception is caught and execution can continue.

Calling iter on an iterable multiple times returns a new iterator each time with distinct states (otherwise, you'd never be able to iterate through a iterable more than once). You can also call iter on the iterator itself, which will just return the same iterator without changing its state. However, note that you cannot call next directly on an iterable.

Let's see the iter and next functions in action with an iterable we're already familiar with -- a list.

>>> lst = [1, 2, 3, 4]
>>> next(lst)             # Calling next on an iterable
TypeError: 'list' object is not an iterator
>>> list_iter = iter(lst) # Creates an iterator for the list
>>> list_iter
<list_iterator object ...>
>>> next(list_iter)       # Calling next on an iterator
1
>>> next(list_iter)       # Calling next on the same iterator
2
>>> next(iter(list_iter)) # Calling iter on an iterator returns itself
3
>>> list_iter2 = iter(lst)
>>> next(list_iter2)      # Second iterator has new state
1
>>> next(list_iter)       # First iterator is unaffected by second iterator
4
>>> next(list_iter)       # No elements left!
StopIteration
>>> lst                   # Original iterable is unaffected
[1, 2, 3, 4]

Since you can call iter on iterators, this tells us that that they are also iterables! Note that while all iterators are iterables, the converse is not true - that is, not all iterables are iterators. You can use iterators wherever you can use iterables, but note that since iterators keep their state, they're only good to iterate through an iterable once:

>>> list_iter = iter([4, 3, 2, 1])
>>> for e in list_iter:
...     print(e)
4
3
2
1
>>> for e in list_iter:
...     print(e)

Analogy: An iterable is like a book (one can flip through the pages) and an iterator for a book would be 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. Calling next on the bookmark moves it to the next page, but does not change the pages in the book. Calling next on the book wouldn't make sense semantically. We can also have multiple bookmarks, all independent of each other.

Iterable Uses

We know that lists are one type of built-in iterable objects. You may have also encountered the range(start, end) function, which creates an iterable of ascending integers from start (inclusive) to end (exclusive).

>>> for x in range(2, 6):
...     print(x)
...
2
3
4
5

Ranges are useful for many things, including performing some operations for a particular number of iterations or iterating through the indices of a list.

There are also some built-in functions that take in iterables and return useful results:

• map(f, iterable) - Creates iterator over f(x) for each x in iterable
• filter(f, iterable) - Creates iterator over x for each x in iterable if f(x)
• zip(iter1, iter2) - Creates iterator over co-indexed pairs (x, y) from both input iterables
• reversed(iterable) - Creates iterator over all the elements in the input iterable in reverse order
• list(iterable) - Creates a list containing all the elements in the input iterable
• tuple(iterable) - Creates a tuple containing all the elements in the input iterable
• sorted(iterable) - Creates a sorted list containing all the elements in the input iterable

Generators

We can create our own custom iterators by writing a generator function, which returns a special type of iterator called a generator. Generator functions have yield statements within the body of the function instead of return statements. Calling a generator function will return a generator object and will not execute the body of the function.

For example, let's consider the following generator function:

def countdown(n):
    print("Beginning countdown!")
    while n >= 0:
        yield n
        n -= 1
    print("Blastoff!")

Calling countdown(k) will return a generator object that counts down from k to 0. Since generators are iterators, we can call iter on the resulting object, which will simply return the same object. Note that the body is not executed at this point; nothing is printed and no numbers are output.

>>> c = countdown(5)
>>> c
<generator object countdown ...>
>>> c is iter(c)
True

So how is the counting done? Again, since generators are iterators, we call next on them to get the next element! The first time next is called, execution begins at the first line of the function body and continues until the yield statement is reached. The result of evaluating the expression in the yield statement is returned. The following interactive session continues from the one above.

>>> next(c)
Beginning countdown!
5

Unlike functions we've seen before in this course, generator functions can remember their state. On any consecutive calls to next, execution picks up from the line after the yield statement that was previously executed. Like the first call to next, execution will continue until the next yield statement is reached. Note that because of this, Beginning countdown! doesn't get printed again.

>>> next(c)
4
>>> next(c)
3

The next 3 calls to next will continue to yield consecutive descending integers until 0. On the following call, a StopIteration error will be raised because there are no more values to yield (i.e. the end of the function body was reached before hitting a yield statement).

>>> next(c)
2
>>> next(c)
1
>>> next(c)
0
>>> next(c)
Blastoff!
StopIteration

Separate calls to countdown will create distinct generator objects with their own state. Usually, generators shouldn't restart. If you'd like to reset the sequence, create another generator object by calling the generator function again.

>>> c1, c2 = countdown(5), countdown(5)
>>> c1 is c2
False
>>> next(c1)
5
>>> next(c2)
5

Here is a summary of the above:

• A generator function has a yield statement and returns a generator object.
• Calling the iter function on a generator object returns the same object without modifying its current state.
• The body of a generator function is not evaluated until next is called on a resulting generator object. Calling the next function on a generator object computes and returns the next object in its sequence. If the sequence is exhausted, StopIteration is raised.

• A generator "remembers" its state for the next next call. Therefore,

  • the first next call works like this:

    1. Enter the function and 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 the yield statement that was previously executed, 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.
• Calling a generator function returns a brand new generator object (like calling iter on an iterable object).
• A generator should not restart unless it's defined that way. To start over from the first element in a generator, just call the generator function again to create a new generator.

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.

>>> def gen_list(lst):
...     yield from lst
...
>>> g = gen_list([1, 2, 3, 4])
>>> next(g)
1
>>> next(g)
2
>>> next(g)
3
>>> next(g)
4
>>> next(g)
StopIteration

Required Questions

WWPD

Q1: WWPD: Iterators

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

python3 ok -q iterators -u

Enter Error if you believe an error occurs, StopIteration if a StopIteration exception is raised, and Iterator if the output is a iterator object.

>>> s = [1, 2, 3, 4]
>>> t = iter(s)
>>> next(s)
______
TypeError
>>> next(t)
______
1
>>> next(t)
______
2
>>> iter(s)
______
Iterator
>>> next(iter(s))
______
1
>>> next(iter(t))
______
3
>>> next(iter(s))
______
1
>>> next(iter(t))
______
4
>>> next(t)
______
StopIteration

Toggle Solution

>>> r = range(6)
>>> r_iter = iter(r)
>>> next(r_iter)
______
0
>>> [x + 1 for x in r]
______
[1, 2, 3, 4, 5, 6]
>>> [x + 1 for x in r_iter]
______
[2, 3, 4, 5, 6]
>>> next(r_iter)
______
StopIteration
>>> list(range(-2, 4))   # Converts an iterable into a list
______
[-2, -1, 0, 1, 2, 3]

Toggle Solution

>>> map_iter = map(lambda x : x + 10, range(5))
>>> next(map_iter)
______
10
>>> next(map_iter)
______
11
>>> list(map_iter)
______
[12, 13, 14]
>>> for e in filter(lambda x : x % 2 == 0, range(1000, 1008)):
...     print(e)
...
______
1000
1002
1004
1006
>>> [x + y for x, y in zip([1, 2, 3], [4, 5, 6])]
______
[5, 7, 9]
>>> for e in zip([10, 9, 8], range(3)):
...   print(tuple(map(lambda x: x + 2, e)))
...
______
(12, 2)
(11, 3)
(10, 4)

Toggle Solution

Q2: WWPD: Generators

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

python3 ok -q generators -u

Enter Error if you believe an error occurs, Function if the output is a function object, and Generator if the output is a generator object.

def gen():
    print("Starting here")
    i = 0
    while i < 6:
        print("Before yield")
        yield i
        print("After yield")
        i += 1

>>> next(gen)
______
TypeError
>>> gen
______
<function gen ...>
>>> g = gen()
>>> g
______
<generator object gen ...>
>>> g == iter(g)
______
True
>>> next(g)
______
Starting here
Before yield
0
>>> next(g)
______
After yield
Before yield
1
>>> next(g)
______
After yield
Before yield
2
>>> g2 = gen()
>>> next(g2)
______
Starting here
Before yield
0
>>> iter(g2)
______
<generator object gen ...>
>>> next(iter(g))
______
After yield
Before yield
3
>>> next(gen())
______
Starting here
Before yield
0

Toggle Solution

Coding Practice

Q3: Scale

Implement the generator function scale(s, k), which yields elements of the given iterable s, scaled by k. As an extra challenge, try writing this function using a yield from statement!

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

# Alternate solution
def scale(s, k):
    yield from map(lambda x: x*k, s)

Toggle Solution

Use Ok to test your code:

python3 ok -q scale

Q4: 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. You may assume that s has at least k values.

To raise an exception, use a raise statement:

>>> def foo():
...     raise ValueError("This is an error message.")
...
>>> foo()
ValueError: This is an error message.

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)
    3
    >>> next(t)
    2
    >>> next(t)
    ValueError
    >>> list(trap(range(5), 5))
    ValueError
    >>> t2 = trap(map(abs, reversed(range(-6, -4))), 2)
    >>> next(t2)
    5
    >>> next(t2)
    6
    >>> next(t2)
    ValueError
    """
    "*** YOUR CODE HERE ***"
    t = iter(s)
    for _ in range(k):
        yield next(t)
    raise ValueError("It's a trap!")

Video walkthrough: https://youtu.be/TRqNVgQSScI?t=57m2s

Toggle Solution

Use Ok to test your code:

python3 ok -q trap

Optional Questions

Q5: Hailstone

Write a generator that outputs the hailstone sequence from homework 1.

Here's a quick reminder 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.

For some extra practice, try writing a solution using recursion. Since hailstone returns a generator, you can yield from a call to hailstone!

def hailstone(n):
    """
    >>> for num in hailstone(10):
    ...     print(num)
    ...
    10
    5
    16
    8
    4
    2
    1
    """
    "*** YOUR CODE HERE ***"
    while n > 1:
        yield n
        if n % 2 == 0:
            n //= 2
        else:
            n = n * 3 + 1
    yield n

# Alternate Solution

def hailstone_alt(n):
    yield n
    if n > 1:
        if n % 2 == 0:
            yield from hailstone_alt(n // 2)
        else:
            yield from hailstone_alt(n * 3 + 1)

Video walkthrough: https://youtu.be/fQlIJa2_yqw?t=1h18m52s

Toggle Solution

Use Ok to test your code:

python3 ok -q hailstone

Q6: 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.

    >>> repeated([10, 9, 10, 9, 9, 10, 8, 8, 8, 7], 2)
    9
    >>> repeated([10, 9, 10, 9, 9, 10, 8, 8, 8, 7], 3)
    8
    >>> s = [3, 2, 1, 2, 1, 4, 4, 5, 5, 5]
    >>> repeated(trap(s, 7), 2)
    4
    >>> repeated(trap(s, 10), 3)
    5
    >>> print(repeated([4, None, None, None], 3))
    None
    """
    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

Video walkthrough: https://youtu.be/fQlIJa2_yqw?t=1h5m35s

Toggle Solution

Use Ok to test your code:

python3 ok -q repeated

Q7: Merge

Implement merge(s0, s1), which takes two iterables s0 and s1 whose elements are ordered. merge yields elements from s0 and s1 in sorted order, eliminating repetition. You may assume s0 and s1 themselves do not contain repeats, and that none of the elements of either are None. You may not assume that the iterables are finite; either may produce an infinite stream of results.

You will probably find it helpful to use the two-argument version of the built-in next function: next(s, v) is the same as next(s), except that instead of raising StopIteration when s runs out of elements, it returns v.

See the doctest for examples of behavior.

def merge(s0, s1):
    """Yield the elements of strictly increasing iterables s0 and s1, removing
    repeats. Assume that s0 and s1 have no repeats. s0 or s1 may be infinite
    sequences.

    >>> m = merge([0, 2, 4, 6, 8, 10, 12, 14], [0, 3, 6, 9, 12, 15])
    >>> type(m)
    <class 'generator'>
    >>> list(m)
    [0, 2, 3, 4, 6, 8, 9, 10, 12, 14, 15]
    >>> def big(n):
    ...    k = 0
    ...    while True: yield k; k += n
    >>> m = merge(big(2), big(3))
    >>> [next(m) for _ in range(11)]
    [0, 2, 3, 4, 6, 8, 9, 10, 12, 14, 15]
    """
    i0, i1 = iter(s0), iter(s1)
    e0, e1 = next(i0, None), next(i1, None)
    "*** YOUR CODE HERE ***"
    while True:
        if e0 is None and e1 is None:
            raise StopIteration
        elif e0 is None or e1 is not None and e1 < e0:
            yield e1
            e1 = next(i1, None)
        elif e1 is None or e0 is not None and e0 < e1:
            yield e0
            e0 = next(i0, None)
        else:
            yield e0
            e0, e1 = next(i0, None), next(i1, None)

Toggle Solution

Use Ok to test your code:

python3 ok -q merge

Q8: 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 is a generator of natural numbers with remainder 1 when divided by m. The last generator yields natural numbers with remainder m - 1 when divided by m.

Hint: You can call the naturals function to create a generator of infinite natural numbers.

Hint: Consider defining an inner generator function. Each yielded generator varies only in that the elements of each generator have a particular remainder when divided by m. What does that tell you about the argument(s) that the inner function should take in?

def remainders_generator(m):
    """
    Yields m generators. The ith yielded generator yields natural numbers whose
    remainder is i when divided by m.

    >>> import types
    >>> [isinstance(gen, types.GeneratorType) for gen in remainders_generator(5)]
    [True, True, True, True, True]
    >>> remainders_four = remainders_generator(4)
    >>> for i in range(4):
    ...     print("First 3 natural numbers with remainder {0} when divided by 4:".format(i))
    ...     gen = next(remainders_four)
    ...     for _ in range(3):
    ...         print(next(gen))
    First 3 natural numbers with remainder 0 when divided by 4:
    4
    8
    12
    First 3 natural numbers with remainder 1 when divided by 4:
    1
    5
    9
    First 3 natural numbers with remainder 2 when divided by 4:
    2
    6
    10
    First 3 natural numbers with remainder 3 when divided by 4:
    3
    7
    11
    """
    "*** YOUR CODE HERE ***"
    def gen(i):
        for e in naturals():
            if e % m == i:
                yield e
    for i in range(m):
        yield gen(i)

Toggle Solution

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.

Use Ok to test your code:

python3 ok -q remainders_generator

Q9: 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_generator(*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_generator([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]

Toggle Solution

Use Ok to test your code:

python3 ok -q zip_generator