Lab 6: Mutability, Iterators
Due by 11:59pm on Wednesday, October 4.
Starter Files
Download lab06.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the Ok autograder.
Required Questions
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Mutability
Consult the drop-down if you need a refresher on mutability. It's okay to skip directly to the questions and refer back here should you get stuck.
Some objects in Python, such as lists and dictionaries, are mutable, meaning that their contents or state can be changed. Other objects, such as numeric types, tuples, and strings, are immutable, meaning they cannot be changed once they are created.
Let's imagine you order a mushroom and cheese pizza from La Val's, and they represent your order as a list:
>>> pizza = ['cheese', 'mushrooms']
With list mutation, they can update your order by mutating pizza
directly
rather than having to create a new list:
>>> pizza.append('onions')
>>> pizza
['cheese', 'mushrooms', 'onions']
Aside from append
, there are various other list mutation methods:
append(elem)
: Addelem
to the end of the list. ReturnNone
.extend(s)
: Add all elements of iterables
to the end of the list. ReturnNone
.insert(i, elem)
: Insertelem
at indexi
. Ifi
is greater than or equal to the length of the list, thenelem
is inserted at the end. This does not replace any existing elements, but only adds the new elementelem
. ReturnNone
.remove(elem)
: Remove the first occurrence ofelem
in list. Errors ifelem
is not in the list. ReturnNone
otherwise.pop(i)
: Remove and return the element at indexi
.pop()
: Remove and return the last element.
An item assignment statement also changes an element of a list. For example:
>>> pizza
['cheese', 'mushrooms', 'onions']
>>> pizza[1] = 'tomatoes'
>>> pizza
['cheese', 'tomatoes', 'onions']
Q1: WWPD: List-Mutation
Important: For all WWPD questions, type
Function
if you believe the answer is<function...>
,Error
if it errors, andNothing
if nothing is displayed.
Use Ok to test your knowledge with the following "What Would Python Display?" questions:
python3 ok -q list-mutation -u
>>> s = [6, 7, 8]
>>> print(s.append(6))
______None
>>> s
______[6, 7, 8, 6]
>>> s.insert(0, 9)
>>> s
______[9, 6, 7, 8, 6]
>>> x = s.pop(1)
>>> s
______[9, 7, 8, 6]
>>> s.remove(x)
>>> s
______[9, 7, 8]
>>> a, b = s, s[:]
>>> a is s
______True
>>> b == s
______True
>>> b is s
______False
>>> s = [3]
>>> s.extend([4, 5])
>>> s
______[3, 4, 5]
>>> s.extend([s.append(9), s.append(10)])
>>> s
______[3, 4, 5, 9, 10, None, None]
Q2: Insert Items
Write a function which takes in a list s
, a value before
, and a value
after
. It inserts after
just after each value equal to before
in s
. It
returns s
.
Important: No new lists should be created or returned.
Note: If the values passed into
before
andafter
are equal, make sure you're not creating an infinitely long list while iterating through it. If you find that your code is taking more than a few seconds to run, the function may be in an infinite loop of inserting new values.
def insert_items(s, before, after):
"""Insert after into s after each occurrence of before and then return s.
>>> test_s = [1, 5, 8, 5, 2, 3]
>>> new_s = insert_items(test_s, 5, 7)
>>> new_s
[1, 5, 7, 8, 5, 7, 2, 3]
>>> test_s
[1, 5, 7, 8, 5, 7, 2, 3]
>>> new_s is test_s
True
>>> double_s = [1, 2, 1, 2, 3, 3]
>>> double_s = insert_items(double_s, 3, 4)
>>> double_s
[1, 2, 1, 2, 3, 4, 3, 4]
>>> large_s = [1, 4, 8]
>>> large_s2 = insert_items(large_s, 4, 4)
>>> large_s2
[1, 4, 4, 8]
>>> large_s3 = insert_items(large_s2, 4, 6)
>>> large_s3
[1, 4, 6, 4, 6, 8]
>>> large_s3 is large_s
True
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q insert_items
Iterators
Consult the drop-down if you need a refresher on iterators. It's okay to skip directly to the questions and refer back here should you get stuck.
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 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, aStopIteration
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).
However, note that you cannot call next
directly on an iterable.
Calling iter
on an iterator simply returns the iterator and is thus not useful.
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 callingiter
on a bookmark gives you the bookmark itself, without changing its position at all. Callingnext
on the bookmark moves it to the next page, but does not change the pages in the book. Callingnext
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 an iterator overf(x)
forx
initerable
. In some cases, computing a list of the values in this iterable will give us the same result as [func(x)
forx
initerable
]. However, it's important to keep in mind that iterators can potentially have infinite values because they are evaluated lazily, while lists cannot have infinite elements.filter(f, iterable)
- Creates an iterator overx
for eachx
initerable
iff(x)
zip(iterables*)
- Creates an iterator over co-indexed tuples with elements from each of theiterables
reversed(iterable)
- Creates an iterator over all the elements in the input iterable in reverse orderlist(iterable)
- Creates a list containing all the elements in the inputiterable
tuple(iterable)
- Creates a tuple containing all the elements in the inputiterable
sorted(iterable)
- Creates a sorted list containing all the elements in the inputiterable
reduce(f, iterable)
- Must be imported withfunctools
. Apply function of two argumentsf
cumulatively to the items ofiterable
, from left to right, so as to reduce the sequence to a single value.
Q3: WWPD: Iterators
Important: Enter
StopIteration
if aStopIteration
exception occurs,Error
if you believe a different error occurs, andIterator
if the output is an iterator object.
Use Ok to test your knowledge with the following "What Would Python Display?" questions:
python3 ok -q iterators-wwpd -u
Python's built-in
map
,filter
, andzip
functions return iterators, not lists. These built-in functions are different from themy_map
andmy_filter
functions we implemented in Lab 04.
>>> s = [1, 2, 3, 4]
>>> t = iter(s)
>>> next(s)
______Error
>>> next(t)
______1
>>> next(t)
______2
>>> iter(s)
______Iterator
>>> next(iter(s))
______1
>>> next(iter(s))
______1
>>> u = t
>>> next(u)
______3
>>> next(t)
______4
>>> next(t)
______StopIteration
>>> 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
>>> 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 % 4 == 0, range(1000, 1008)):
... print(e)
______1000
1004
>>> [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)
Q4: Count Occurrences
Implement count_occurrences
, which takes an iterator t
and a value x
. It
returns the number of elements equal to x
that appear in the first n
elements of t
.
Important: Call next
on t
exactly n
times. Assume there are at least
n
elements in t
.
Hint: When the same iterator is passed into a function a second time, it should pick up where it left off after the first call, as with
s
from the doctest below.
def count_occurrences(t, n, x):
"""Return the number of times that x is equal to one of the
first n elements of iterator t.
>>> s = iter([10, 9, 10, 9, 9, 10, 8, 8, 8, 7])
>>> count_occurrences(s, 10, 9)
3
>>> s2 = iter([10, 9, 10, 9, 9, 10, 8, 8, 8, 7])
>>> count_occurrences(s2, 3, 10)
2
>>> s = iter([3, 2, 2, 2, 1, 2, 1, 4, 4, 5, 5, 5])
>>> count_occurrences(s, 1, 3) # Only iterate over 3
1
>>> count_occurrences(s, 3, 2) # Only iterate over 2, 2, 2
3
>>> list(s) # Ensure that the iterator has advanced the right amount
[1, 2, 1, 4, 4, 5, 5, 5]
>>> s2 = iter([4, 1, 6, 6, 7, 7, 6, 6, 2, 2, 2, 5])
>>> count_occurrences(s2, 6, 6)
2
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q count_occurrences
Q5: Repeated
Implement repeated
, which takes in an iterator t
and an integer k
greater
than 1. It returns the first value in t
that appears k
times in a row.
Important: Call next
on t
only the minimum number of times required.
Assume that there is an element of t
repeated at least k
times in a row.
Hint: If you are receiving a
StopIteration
exception, yourrepeated
function is likely not identifying the correct value.
def repeated(t, k):
"""Return the first value in iterator t that appears k times in a row,
calling next on t as few times as possible.
>>> s = iter([10, 9, 10, 9, 9, 10, 8, 8, 8, 7])
>>> 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, 1, 2, 1, 4, 4, 5, 5, 5])
>>> 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 ***"
Use Ok to test your code:
python3 ok -q repeated
Check Your Score Locally
You can locally check your score on each question of this assignment by running
python3 ok --score
This does NOT submit the assignment! When you are satisfied with your score, submit the assignment to Gradescope to receive credit for it.
Submit
Make sure to submit this assignment by uploading any files you've edited to the appropriate Gradescope assignment. For a refresher on how to do this, refer to Lab 00.
Optional Questions
These questions are optional, but you must complete them in order to be checked off before the end of the lab period. They are also useful practice!
Q6: Partial Reverse
When working with lists, it is often useful to reverse the list. For example,
reversing the list [1, 2, 3, 4, 5]
will give [5, 4, 3, 2, 1]
. However, in
some situations, it may be more useful to only partially reverse the list and
keep some of its elements in the same order. For example, partially reversing
the list [1, 2, 3, 4, 5]
starting from index 2 until the end of the list will
give [1, 2, 5, 4, 3]
.
Implement the function partial_reverse
which reverses a list starting
from start
until the end of the list. This reversal should be
in-place, meaning that the original list is modified. Do not create a
new list inside your function, even if you do not return it.
The partial_reverse
function returns None
.
def partial_reverse(s, start):
"""Reverse part of a list in-place, starting with start up to the end of
the list.
>>> a = [1, 2, 3, 4, 5, 6, 7]
>>> partial_reverse(a, 2)
>>> a
[1, 2, 7, 6, 5, 4, 3]
>>> partial_reverse(a, 5)
>>> a
[1, 2, 7, 6, 5, 3, 4]
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q partial_reverse