Lab 5: Mutability, Iterators, Generators
Due by 11:59pm on Thursday, July 9.
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Mutability
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.
The two most common mutation operations for lists are item assignment and the
append method.
>>> s = [1, 3, 4]
>>> t = s # A second name for the same list
>>> t[0] = 2 # this changes the first element of the list to 2, affecting both s and t
>>> s
[2, 3, 4]
>>> s.append(5) # this adds 5 to the end of the list, affecting both s and t
>>> t
[2, 3, 4, 5]
There are many other list mutation methods:
append(elem): Addelemto the end of the list. ReturnNone.extend(s): Add all elements of iterablesto the end of the list. ReturnNone.insert(i, elem): Insertelemat indexi. Ifiis greater than or equal to the length of the list, thenelemis inserted at the end. This does not replace any existing elements, but only adds the new elementelem. ReturnNone.remove(elem): Remove the first occurrence ofelemin list. ReturnNone. Errors ifelemis not in the list.pop(i): Remove and return the element at indexi.pop(): Remove and return the last element.
Dictionaries also have item assignment (often used) and pop (rarely used).
>>> d = {2: 3, 4: 16}
>>> d[2] = 4
>>> d[3] = 9
>>> d
{2: 4, 4: 16, 3: 9}
>>> d.pop(4)
16
>>> d
{2: 4, 3: 9}
Iterators
An iterable is any value 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 statement:
for elem in iterable:
# do something
In general, an iterable is an object on which calling the built-in iter
function returns an iterator. An iterator is an object on which calling
the built-in next function returns the next value.
For example, a list is an iterable value.
>>> s = [1, 2, 3, 4]
>>> next(s) # s is iterable, but not an iterator
TypeError: 'list' object is not an iterator
>>> t = iter(s) # Creates an iterator
>>> t
<list_iterator object ...>
>>> next(t) # Calling next on an iterator
1
>>> next(t) # Calling next on the same iterator
2
>>> next(iter(t)) # Calling iter on an iterator returns itself
3
>>> t2 = iter(s)
>>> next(t2) # Second iterator starts at the beginning of s
1
>>> next(t) # First iterator is unaffected by second iterator
4
>>> next(t) # No elements left!
StopIteration
>>> s # Original iterable is unaffected
[1, 2, 3, 4]
You can also use an iterator in a for statement because all iterators are
iterable. But note that since iterators keep their state, they're
only good to iterate through an iterable once:
>>> t = iter([4, 3, 2, 1])
>>> for e in t:
... print(e)
4
3
2
1
>>> for e in t:
... print(e)
There are built-in functions that return iterators. These built-in Python sequence operations are said to compute results lazily.
>>> m = map(lambda x: x * x, [3, 4, 5])
>>> next(m)
9
>>> next(m)
16
>>> f = filter(lambda x: x > 3, [3, 4, 5])
>>> next(f)
4
>>> next(f)
5
>>> z = zip([30, 40, 50], [3, 4, 5])
>>> next(z)
(30, 3)
>>> next(z)
(40, 4)
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 outputted.
>>> 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
yieldstatement and returns a generator object. - Calling the
iterfunction on a generator object returns the same object without modifying its current state. - The body of a generator function is not evaluated until
nextis called on a resulting generator object. Calling thenextfunction on a generator object computes and returns the next object in its sequence. If the sequence is exhausted,StopIterationis raised. A generator "remembers" its state for the next
nextcall. Therefore,the first
nextcall works like this:- Enter the function and run until the line with
yield. - Return the value in the
yieldstatement, but remember the state of the function for futurenextcalls.
- Enter the function and run until the line with
And subsequent
nextcalls work like this:- Re-enter the function, start at the line after the
yieldstatement that was previously executed, and run until the nextyieldstatement. - Return the value in the
yieldstatement, but remember the state of the function for futurenextcalls.
- Re-enter the function, start at the line after the
- Calling a generator function returns a brand new generator object (like
calling
iteron 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. 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
Mutability
Q1: WWPD: List-Mutation
Important: For all WWPD questions, type
Functionif you believe the answer is<function...>,Errorif it errors, andNothingif 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
>>> a.pop()
______8
>>> a + b
______[9, 7, 9, 7, 8]
>>> s = [3]
>>> s.extend([4, 5])
>>> s
______[3, 4, 5]
>>> a
______[9, 7]
>>> s.extend([s.append(9), s.append(10)])
>>> s
______[3, 4, 5, 9, 10, None, None]
Q2: Insert Items
Write a function that takes in a list s, a value before, and a value
after. It modifies s in place by inserting after just after each
value equal to before in s. It returns s.
Important: No new lists should be created.
Note: If the values passed into
beforeandafterare 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: list[int], before: int, after: int) -> list[int]:
"""Insert after into s following 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
Q3: Group By
Write a function that takes in a list s and a function fn, and returns a dictionary that groups the elements of s based on the result of applying fn.
- The dictionary should have one key for each unique result of applying
fnto elements ofs. - The value for each key should be a list of all elements in
sthat, when passed tofn, produce that key (what it evaluates to).
In other words, for each element e in s, determine fn(e) and add e to the list corresponding to fn(e) in the dictionary.
def group_by(s: list[int], fn) -> dict[int, list[int]]:
"""Return a dictionary of lists that together contain the elements of s.
The key for each list is the value that fn returns when called on any of the
values of that list.
>>> group_by([12, 23, 14, 45], lambda p: p // 10)
{1: [12, 14], 2: [23], 4: [45]}
>>> group_by(range(-3, 4), lambda x: x * x)
{9: [-3, 3], 4: [-2, 2], 1: [-1, 1], 0: [0]}
"""
grouped = {}
for ____ in ____:
key = ____
if key in grouped:
____
else:
grouped[key] = ____
return grouped
Use Ok to test your code:
python3 ok -q group_by
Iterators
Q4: WWPD: Iterators
Important: Enter
StopIterationif aStopIterationexception occurs,Errorif you believe a different error occurs, andIteratorif the output is an iterator object.
Important: Python's built-in function
map,filter, andzipreturn iterators, not lists.
Use Ok to test your knowledge with the following "What Would Python Display?" questions:
python3 ok -q iterators-wwpd -u
>>> s = [1, 2, 3, 4]
>>> t = iter(s)
>>> next(s)
______Error
>>> next(t)
______1
>>> next(t)
______2
>>> next(iter(s))
______1
>>> next(iter(s))
______1
>>> u = t
>>> next(u)
______3
>>> next(t)
______4
>>> 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]
Q5: Count Occurrences
Implement count_occurrences, which takes an iterator t, an integer n, and
a value x. It returns the number of elements in the
first n elements of t that are equal to x.
You can assume that t has at least n elements.
Important: You should call
nextontexactlyntimes. If you need to iterate through more thannelements, think about how you can optimize your solution.
def count_occurrences(t: Iterator[int], n: int, x: int) -> int:
"""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
>>> t = iter([10, 9, 10, 9, 9, 10, 8, 8, 8, 7])
>>> count_occurrences(t, 3, 10)
2
>>> u = iter([3, 2, 2, 2, 1, 2, 1, 4, 4, 5, 5, 5])
>>> count_occurrences(u, 1, 3) # Only iterate over 3
1
>>> count_occurrences(u, 3, 2) # Only iterate over 2, 2, 2
3
>>> list(u) # Ensure that the iterator has advanced the right amount
[1, 2, 1, 4, 4, 5, 5, 5]
>>> v = iter([4, 1, 6, 6, 7, 7, 6, 6, 2, 2, 2, 5])
>>> count_occurrences(v, 6, 6)
2
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q count_occurrences
Generators
Q6: Generate Permutations
Given a sequence of unique elements, a permutation of the sequence is a list
containing the elements of the sequence in some arbitrary order. For example,
[1, 2, 3], [2, 1, 3], [1, 3, 2], and [3, 2, 1] are some of the
permutations of the sequence [1, 2, 3].
Implement perms, a generator function that takes in a sequence seq
and returns a generator that yields all permutations of seq. For this question,
assume that seq will not be empty.
Permutations may be yielded in any order.
Hint: Remember, it's possible to loop over generator objects because generators are iterators!
def perms(seq):
"""Generates all permutations of the given sequence. Each permutation is a
list of the elements in SEQ in a different order. The permutations may be
yielded in any order.
>>> p = perms([100])
>>> type(p)
<class 'generator'>
>>> next(p)
[100]
>>> try: # Prints "No more permutations!" if calling next would cause an error
... next(p)
... except StopIteration:
... print('No more permutations!')
No more permutations!
>>> sorted(perms([1, 2, 3])) # Returns a sorted list containing elements of the generator
[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]
>>> sorted(perms((10, 20, 30)))
[[10, 20, 30], [10, 30, 20], [20, 10, 30], [20, 30, 10], [30, 10, 20], [30, 20, 10]]
>>> sorted(perms("ab"))
[['a', 'b'], ['b', 'a']]
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q perms
Check Your Score Locally
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python3 ok --score
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Optional Questions
These questions are optional. If you don't complete them, you will still receive credit for this assignment. They are great practice, so do them anyway!
Q7: 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.
You may assume that there is an element of t repeated at least k times in a row.
Important: Call
nextontonly the minimum number of times required. If you are receiving aStopIterationexception, yourrepeatedfunction is callingnexttoo many times.
def repeated(t: Iterator[int], k: int) -> int:
"""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
>>> t = iter([10, 9, 10, 9, 9, 10, 8, 8, 8, 7])
>>> repeated(t, 3)
8
>>> u = iter([3, 2, 2, 2, 1, 2, 1, 4, 4, 5, 5, 5])
>>> repeated(u, 3)
2
>>> repeated(u, 3)
5
>>> v = iter([4, 1, 6, 6, 7, 7, 8, 8, 2, 2, 2, 5])
>>> repeated(v, 3)
2
"""
assert k > 1
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q repeated