Lab 11: Iterators and Generators
Due at 11:59pm on Friday, 11/10/2017.
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.
Check that you have successfully submitted your code on
okpy.org.
- To receive credit for this lab, you must complete Questions 1 - 5 in lab11.py and submit through OK.
- Questions 6 - 10 are extra practice. They can be found in the lab11_extra.py file. It is recommended that you complete these problems on your own 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)
try:
while True:
elem = next(the_iterator)
# do something
except StopIteration:
pass
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
.
Q1: 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)
______1
>>> next(iter(next(t)))
______2
>>> list(t)
______[]
>>> next(map(lambda x: x * x, filter(lambda y: y > 4, range(10))))
______25
>>> r = iter(reversed(range(10000)))
>>> next(r) - next(r)
______1
>>> xs = [2, 3, 4, 5]
>>> y, z = iter(xs), iter(xs)
>>> next(zip(y, z))
______(2, 2)
>>> next(zip(y, y))
______(3, 4)
Generators
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
function makes a generator object return itself without modifying its current state. - Calling the
next
function 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
anditer
explicitly defined, but a generator can be written as a mere function with ayield
in it. next
in an iterator usesreturn
, but a generator usesyield
.A generator "remembers" its state for the next
next
call. Therefore,the first
next
call works like this:- Enter the function, run until the line with
yield
. - Return the value in the
yield
statement, but remember the state of the function for futurenext
calls.
- Enter the function, run until the line with
And subsequent
next
calls work like this:- Re-enter the function, start at the line after
yield
, and run until the nextyield
statement. - Return the value in the
yield
statement, but remember the state of the function for futurenext
calls.
- Re-enter the function, start at the line after
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.
Q2: 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?
>>> 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
Q3: 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
Q4: 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)
...
5
4
3
2
1
0
>>> for number in countdown(2):
... print(number)
...
2
1
0
"""
"*** YOUR CODE HERE ***"
while n >= 0:
yield n
n = n - 1
Use Ok to test your code:
python3 ok -q countdown
Q5: 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.
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)
while k > 0:
yield next(t)
k -= 1
raise ValueError("It's a trap!")
Use Ok to test your code:
python3 ok -q trap
Extra Questions
Q6: Hailstone
Write a generator that outputs the hailstone sequence from homework 1.
Here's a quick reminder of how the hailstone sequence is defined:
- Pick a positive integer
n
as the start. - If
n
is even, divide it by 2. - If
n
is odd, multiply it by 3 and add 1. - 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
Q7: 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)
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
Use Ok to test your code:
python3 ok -q repeated
Q8: 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. You can also assume that s0
and s1 represent 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:
return
elif e0 is None:
yield e1
e1 = next(i1, None)
elif e1 is None:
yield e0
e0 = next(i0, None)
else:
yield min(e0, e1)
if e0 < e1:
e0 = next(i0, None)
elif e1 < e0:
e1 = next(i1, None)
else:
e0, e1 = next(i0, None), next(i1, None)
Use Ok to test your code:
python3 ok -q merge
Q9: Remainder Generator
Like functions, generators can also be higher-order. For this problem, we will be writingremainders_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
Q10: 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]
Use Ok to test your code:
python3 ok -q zip_generator