Lab 11: Iterators and Generators

Due at 11:59pm on 4/15/2016.

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

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 lab11.py and submit through OK.
• Questions 8, 9, 10, 11 are extra practice. They can be found in the lab11_extra.py file. It is recommended that you complete these problems on your on time.

Helpful Hints

We have a new feature we are testing in OK. This feature tries to give helpful hints when it sees you struggling with unlocking questions.

To see if everything works you will be randomly placed in a group to either get the hints or not. If you wish to make sure the hints appear for you in this lab, run an unlocking command with the flag --guidance. For example if the command is usually python3 ok -q XXX -u instead use python3 ok --guidance -q XXX -u. Once you use this flag, all following unlocking commands will give you helpful hints regardless of if you include the flag again.

Iterables and Iterators

Remember the for loop? (We really hope so!)

for elem in something_iterable:
    # do something

for loops only work with iterables, which we can define as values that are acceptable to the built-in iter function. The for loop above, for example, behaves like this:

the_iterator = iter(something_iterable)
try:
    while True:
        elem = the_iterator.__next__()
        # do something
except StopIteration:
    pass

That is, it first calls iter to create an iterator, saving it in some new, hidden variable (we've called it the_iterator here). It then repeatedly calls the __next__() method on this iterator to get values of elem and stops when that method raises StopIteration.
The term iterator simply means "an object that has a __next__() method that normally either returns a value or raises StopIteration to indicate completion of the iteration."

For something_iterable to be acceptable to the iter function, either it must

1. define an __iter__ method that returns an iterator, as defined above, or
2. define a __getitem__ method that takes an integer index argument and either returns a value or raises IndexError when the index argument exceeds some bound (such as the length of a list).

iter handles these two cases as follows:

1. In the first case, iter simply returns the result of applying __iter__.
2. In the second, it creates an iterator object whose __next__ method calls __getitem__(0), __getitem__(1), ..., until it receives IndexError, at which point this manufactured iterator raises StopIteration.

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

The iterators produced by the predefined functions in the Python library implement __iter__ methods that just return themselves (iterators you define yourself should do this also). This is convenient, because it means that if you have an expression that produces an iterator (such as calls on the map and zip builtin functions), you can do a for loop over it as well. But be careful. Iterators are not the same as familiar iterables like lists or tuples. Compare the following two cases:

>>> L = [ 1, 2 ]
>>> I = iter(L)    # I is an iterator over L
>>> for x in L: print x
1
2
>>> for x in L: print x
1
2
>>> for x in I: print x
1
2
>>> for x in I: print x

The last for loop produces nothing. Why? Because I is an iterator: once its __next__ method raises StopIteration, it produces no more values. Since I.__iter__() simply returns I, therefore, the last for loop finds no more values, whereas the second for x in L loop calls __iter__ on L, which produces a new iterator. This is why we consider the terminology somewhat confusing: lists and iterators may both be iterables and both be useable in for, but as the example shows, they behave differently.

Here is a table summarizing the required methods on iterables and iterators. Python also has more documentation about iterator types.

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 Skipper:
    def __init__(self, seq):
        self.seq = seq
        self.step = 1
        self.current = 0

    def __next__(self):
        if self.current > len(self.seq):
            raise StopIteration
        val = self.seq[self.current]
        self.current += self.step
        self.step += 1
        return val

    def __iter__(self):
        return self

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

>>> for num in Skipper([3, 4, 5, 6, 7, 8, 9, 10]):
...     print(num)
3
4
6
9

That is, we skip through the elements of some sequence in ever-larger steps.

We can also use Skipper to create a new kind of iterable list:

class SkippingList(list):

    def __iter__(self):
        return Skipper(self)

and now we can write:

>>> aList = SkippingList([3, 4, 5, 6, 7, 8, 9, 10])
>>> for num in aList:
...     print(num)
3
4
6
9

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

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

Try this!

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

Then again:

>>> [num for num in iterator]

The outputs of the list comprehensions are like that 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

Question 3: WWPP: Odds and Evens

So far, the __iter__ method of our iterators only returns self. What if we have called next a few times and then want to start at the beginning? Intuitively we should create a new iterator that would start at the beginning. However, our current iterator implementations won't allow that.

Consider the following OddNaturalsIterator and EvenNaturalsIterator, which implementation allows us to start a new iterator at the beginning?

class OddNaturalsIterator():
  def __init__(self):
    self.current = 1

  def __next__(self):
    result = self.current
    self.current += 2
    return result

  def __iter__(self):
    return self

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

  def __next__(self):
    result = self.current
    self.current += 2
    return result

  def __iter__(self):
    return EvenNaturalsIterator()

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

python3 ok -q odds_evens -u

>>> odds = OddNaturalsIterator()
>>> odd_iter1 = iter(odds)
>>> odd_iter2 = iter(odds)
>>> next(odd_iter1)
______
1
>>> next(odd_iter1)
______
3
>>> next(odd_iter1)
______
5
>>> next(odd_iter2)
______
7
>>> next(odd_iter1)
______
9
>>> next(odd_iter2)
______
11

>>> evens = EvenNaturalsIterator()
>>> even_iter1 = iter(evens)
>>> even_iter2 = iter(evens)    
>>> next(even_iter1)
______
0
>>> next(even_iter1)
______
2
>>> next(even_iter1)
______
4
>>> next(even_iter2)
______
0
>>> next(even_iter2)
______
2

Question 4: 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

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>():
    <variable> = <value>
    while <predicate>:
        yield <value>
        <increment variable>

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.

Question 5: WWPP: Generators

Use OK to test your knowledge with the following What would Python print 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__()
______
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 6: 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 7: Hailstone

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

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

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: 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. You may also assume s0 and s1 represent infinite sequences; that is, their iterators never raise StopIteration.

See the doctests for example 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. You can also assume that s0
    and s1 represent infinite sequences.

    >>> twos = scale(naturals(), 2)
    >>> threes = scale(naturals(), 3)
    >>> m = merge(twos, threes)
    >>> type(m)
    <class 'generator'>
    >>> [next(m) for _ in range(10)]
    [2, 3, 4, 6, 8, 9, 10, 12, 14, 15]
    """
    i0, i1 = iter(s0), iter(s1)
    e0, e1 = next(i0), next(i1)
    "*** YOUR CODE HERE ***"
    while True:
        yield min(e0, e1)
        if e0 < e1:
            e0 = next(i0)
        elif e1 < e0:
            e1 = next(i1)
        else:
            e0, e1 = next(i0), next(i1)

Use OK to test your code:

python3 ok -q merge

Question 10: 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))
    0
    4
    8
    1
    5
    9
    2
    6
    10
    3
    7
    11
    """
    "*** 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 11: 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