Instructions

Please download lab materials lab05.zip from our QQ group if you don't have one.

In this lab, you have two tasks:

  • Think about what would Python display in section 3 and verify your answer with Ok.
  • Complete the required problems described in section 4 and submit your code to our OJ website.

The starter code for these problems is provided in lab05.py.

Submission: As instructed before, you need to submit your work with Ok by python ok --submit. You may submit more than once before the deadline, and your score of this assignment will be the highest one of all your submissions.

Readings: You might find the following reference to the textbook useful:

Review

Consult this section if you need a refresher on the material for this lab. It's okay to skip directly to the next section and refer back here when you get stuck.

Nonlocal

We say that a variable defined in a frame is local to that frame. A variable is nonlocal to a frame if it is defined in the environment that the frame belongs to but not the frame itself, i.e. in its parent or ancestor frame.

So far, we know that we can access variables in parent frames:

def make_adder(x):
    """ Returns a one-argument function that returns the result of
    adding x and its argument. """
    def adder(y):
        return x + y
    return adder

Here, when we call make_adder, we create a function adder that is able to look up the name x in make_adder's frame and use its value.

However, we haven't been able to modify variable in parent frames. Consider the following function:

def make_withdraw(balance):
    """Returns a function which can withdraw
    some amount from balance

    >>> withdraw = make_withdraw(50)
    >>> withdraw(25)
    25
    >>> withdraw(25)
    0
    """
    def withdraw(amount):
        if amount > balance:
            return "Insufficient funds"
        balance = balance - amount
        return balance
    return withdraw

The inner function withdraw attempts to update the variable balance in its parent frame. Running this function's doctests, we find that it causes the following error:

UnboundLocalError: local variable 'balance' referenced before assignment

Why does this happen? When we execute an assignment statement, remember that we are either creating a new binding in our current frame or we are updating an old one in the current frame. For example, the line balance = ... in withdraw, is creating the local variable balance inside withdraw's frame. This assignment statement tells Python to expect a variable called balance inside withdraw's frame, so Python will not look in parent frames for this variable. However, notice that we tried to compute balance - amount before the local variable was created! That's why we get the UnboundLocalError.

To avoid this problem, we introduce the nonlocal keyword. It allows us to update a variable in a parent frame!

Some important things to keep in mind when using nonlocal

  • nonlocal cannot be used with global variables (names defined in the global frame). If no nonlocal variable is found with the given name, a SyntaxError is raised. A name that is already local to a frame cannot be declared as nonlocal.

Consider this improved example:

def make_withdraw(balance):
    """Returns a function which can withdraw
    some amount from balance

    >>> withdraw = make_withdraw(50)
    >>> withdraw(25)
    25
    >>> withdraw(25)
    0
    """
    def withdraw(amount):
        nonlocal balance
        if amount > balance:
            return "Insufficient funds"
        balance = balance - amount
        return balance
    return withdraw

The line nonlocal balance tells Python that balance will not be local to this frame, so it will look for it in parent frames. Now we can update balance without running into problems.

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 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 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(1, 6):
...     print(x)
...
1
2
3
4

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(iter0, 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)
Beginning countdown!
5
>>> next(c2)
Beginning countdown!
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:
      • Enter the function and run until the line with yield.
      • 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:
      • Re-enter the function, start at the line after the yield statement that was previously executed, and run until the next yield statement.
      • 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

What Would Python Display?

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

python ok -q wwpd -u

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

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

>>> next(t)
______

>>> next(t)
______

>>> iter(s)
______

>>> next(iter(s))
______

>>> next(iter(t))
______

>>> next(iter(s))
______

>>> next(iter(t))
______

>>> next(t)
______

>>> r = range(6)
>>> r_iter = iter(r)
>>> next(r_iter)
______

>>> [x + 1 for x in r]
______

>>> [x + 1 for x in r_iter]
______

>>> next(r_iter)
______

>>> list(range(-2, 4))   # Converts an iterable into a list
______

>>> map_iter = map(lambda x : x + 10, range(5))
>>> next(map_iter)
______

>>> next(map_iter)
______

>>> list(map_iter)
______

>>> for e in filter(lambda x : x % 2 == 0, range(1000, 1008)):
...     print(e)
______

>>> [x + y for x, y in zip([1, 2, 3], [4, 5, 6])]
______

>>> for e in zip([10, 9, 8], range(3)):
...   print(tuple(map(lambda x: x + 2, e)))
______

Required Problems

In this section, you are required to complete the problems below and submit your code to OJ website as instructed in lab00 to get your answer scored.

Problem 1: Count (100pts)

Implement count, which takes in an iterator t and returns the number of times the value x appears in the first n elements of t. A value appears in a sequence of elements if it is equal to an entry in the sequence.

Note: You can assume that t will have at least n elements.

def count(t, n, x):
    """Return the number of times that x appears in the first n elements of iterator t.

    >>> s = iter([10, 9, 10, 9, 9, 10, 8, 8, 8, 7])
    >>> count(s, 10, 9)
    3
    >>> s2 = iter([10, 9, 10, 9, 9, 10, 8, 8, 8, 7])
    >>> count(s2, 3, 10)
    2
    >>> s = iter([3, 2, 2, 2, 1, 2, 1, 4, 4, 5, 5, 5])
    >>> count(s, 1, 3)
    1
    >>> count(s, 4, 2)
    3
    >>> next(s)
    2
    >>> s2 = iter([4, 1, 6, 6, 7, 7, 8, 8, 2, 2, 2, 5])
    >>> count(s2, 6, 6)
    2
    """
    "*** YOUR CODE HERE ***"

Problem 2: Repeated (100pts)

Implement repeated, which takes in an iterator t and returns the first value in t that appears k times in a row.

Note: You can assume that the iterator t will have a value that appears at least k times in a row. If you are receiving a StopIteration, your repeated function is likely not identifying the correct value.

Your implementation should iterate through the items in a way such that if the same iterator is passed into repeated twice, it should continue in the second call at the point it left off in the first. An example of this behavior is in the doctests.

def repeated(t, k):
    """Return the first value in iterator T that appears K times in a row.
    Iterate through the items such that if the same iterator is passed into
    the function twice, it continues in the second call at the point it left
    off in the first.

    >>> s = iter([10, 9, 10, 9, 9, 10, 8, 8, 8, 7])
    >>> repeated(s, 1)
    10
    >>> repeated(s, 2)
    9
    >>> s2 = iter([10, 9, 10, 9, 9, 10, 8, 8, 8, 7])
    >>> repeated(s2, 3)
    8
    >>> s = iter([3, 2, 2, 2, 2, 2, 2, 4, 4, 5, 5, 5])
    >>> repeated(s, 3)
    2
    >>> repeated(s, 3)
    2
    >>> repeated(s, 3)
    5
    >>> s2 = iter([4, 1, 6, 6, 7, 7, 8, 8, 2, 2, 2, 5])
    >>> repeated(s2, 3)
    2
    """
    assert k >= 1
    "*** YOUR CODE HERE ***"

Problem 3: Scale (100pts)

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

As an extra challenge, try writing this function using a yield from statement!

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

    >>> m = scale(iter([1, 5, 2]), 5)
    >>> type(m)
    <class 'generator'>
    >>> list(m)
    [5, 25, 10]
    >>> # generators allow us to represent infinite sequences!!!
    >>> def naturals():
    ...     i = 0
    ...     while True:
    ...         yield i
    ...         i += 1
    >>> m = scale(naturals(), 2)
    >>> [next(m) for _ in range(5)]
    [0, 2, 4, 6, 8]
    """
    "*** YOUR CODE HERE ***"

Problem 4: Merge (100pts)

Write a generator function merge that takes in two infinite generators a and b that are in increasing order without duplicates and returns a generator that has all the elements of both generators, in increasing order, without duplicates.

Note: You may assume that both a and b contains unlimited elements. As a result, calling next(a) and next(b) will never throw StopIteration exception.

def merge(a, b):
    """Merge two generators that are in increasing order and without duplicates.
    Return a generator that has all elements of both generators in increasing
    order and without duplicates.

    >>> def sequence(start, step):
    ...     while True:
    ...         yield start
    ...         start += step
    >>> a = sequence(2, 3) # 2, 5, 8, 11, 14, ...
    >>> b = sequence(3, 2) # 3, 5, 7, 9, 11, 13, 15, ...
    >>> result = merge(a, b) # 2, 3, 5, 7, 8, 9, 11, 13, 14, 15
    >>> [next(result) for _ in range(10)]
    [2, 3, 5, 7, 8, 9, 11, 13, 14, 15]
    """
    "*** YOUR CODE HERE ***"

Problem 5: Hailstone (100pts)

Write a generator that outputs the hailstone sequence from hw01.

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):
    """Return a generator that outputs the hailstone sequence.

    >>> for num in hailstone(10):
    ...     print(num)
    10
    5
    16
    8
    4
    2
    1
    """
    "*** YOUR CODE HERE ***"