Lab 6: Nonlocal, Mutability, Iterators and Generators
Due by 11:59pm on Tuesday, March 2.
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.
Topics
Consult this section if you need a refresher on the material for this lab. It's okay to skip directly to the questions and refer back here should 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 function which takes in one argument y, and returns x + y.
>>> add_three = make_adder(3)
>>> add_three(2)
5
"""
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 variables defined in parent frames. Consider the following function:
def make_adder_cumulative(x):
"""
Returns a function which takes in one argument y and returns y + x + the sum of all previous y's.
It also updates x to this new total.
>>> add_all = make_adder_cumulative(3)
>>> add_all(2) # 3 + 2
5
>>> add_all(5) # 3 + 2 + 5
10
"""
def adder(y):
total = x + y
x = total
return total
return adder
Here, when we call make_adder_cumulative
, we create a function adder
that is able to
look up the name x
in make_adder_cumulative
's frame and use its value. However, adder
also attempts to update the variable x
in its parent frame. Running this function's doctests, we
find that it causes the following error:
UnboundLocalError: local variable 'x' referenced before assignment
Why does this happen? When we execute an assignment statement, 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 x = x + y
in adder
,
is creating the local variable x
inside adder
's frame. This
assignment statement tells Python to expect a variable called x
inside
adder
's frame, so Python will not look in parent frames for this variable.
However, notice that we tried to compute x + y
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!
The only exception is the global frame. You cannot update a variable in the
global frame using nonlocal
.
Consider this example:
def make_adder_cumulative(x):
"""
Returns a function which takes in one argument y and returns y + x + the sum of all previous y's.
It also updates x to this new total.
>>> add_all = make_adder_cumulative(3)
>>> add_all(2) # 3 + 2
5
>>> add_all(5) # 3 + 2 + 5
10
"""
def adder(y):
nonlocal x
total = x + y
x = total
return total
return adder
The line nonlocal x
tells Python that x
will not be local to this frame, so it will look for it in parent frames. Now we can update x
without running into problems.
Mutability
We say that an object is mutable if its state can change as code is executed. The process of changing an object's state is called mutation. Examples of mutable objects include lists and dictionaries. Examples of objects that are not mutable include tuples and functions.
We have seen how to use the ==
operator to check if two expressions evaluate
to equal values. We now introduce a new comparison operator, is
, that
checks whether two expressions evaluate to the same values.
Wait, what's the difference? For primitive values, there is none:
>>> 2 + 2 == 3 + 1
True
>>> 2 + 2 is 3 + 1
True
This is because all primitives have the same identity under the hood. However, with non-primitive values, such as lists, each object has its own identity. That means you can construct two objects that may look exactly the same but have different identities.
>>> lst1 = [1, 2, 3, 4]
>>> lst2 = [1, 2, 3, 4]
>>> lst1 == lst2
True
>>> lst1 is lst2
False
Here, although the lists referred to by lst1
and lst2
have equal
contents, they are not the same object. In other words, they are the same in
terms of equality, but not in terms of identity.
This is important in our discussion of mutability because when we mutate an object, we simply change its state, not its identity.
>>> lst1 = [1, 2, 3, 4]
>>> lst2 = lst1
>>> lst1.append(5)
>>> lst2
[1, 2, 3, 4, 5]
>>> lst1 is lst2
True
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 a 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). 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 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 iterable 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 iterator overx
for eachx
initerable
iff(x)
zip(iterables*)
- Creates an iterable over co-indexed tuples with elements from each of theiterables
reversed(iterable)
- Creates 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
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)
5
>>> next(c2)
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 thenext
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 futurenext
calls.
- Enter the function and run until the line with
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 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 the
- 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
Required Questions
Nonlocal WWPD
Q1: WWPD: Nonlocal Quiz
Use Ok to test your knowledge with the following "What Would Python Display?" questions:
python3 ok -q nonlocal_quiz -u
Relevant Topics: Nonlocal
>>> def ba(by):
... def yo(da):
... by += 2
... return by
... return yo(2)
...
>>> ba(3)
>>> def ba(by):
... def yo(da):
... nonlocal by
... by += 2
... return by
... return yo(3)
...
>>> ba(3)
>>> def ba(by):
... def yo(da):
... by.append(da)
... return by
... return yo(5)
...
>>> ba([1, 2, 3])
>>> def ba(by):
... def yo(da):
... yoda = by + da
... return yoda
... return yo(5)
...
>>> ba(5)
Mutability
Q2: List-Mutation
Test your understanding of list mutation with the following questions. What would Python display? Type it in the interpreter if you're stuck!
python3 ok -q list-mutation -u
Note: if nothing would be output by Python, type
Nothing
. If the code would error, typeError
.Relevant Topics: Mutability
>>> lst = [5, 6, 7, 8]
>>> lst.append(6)
______None
>>> lst
______[5, 6, 7, 8, 6]
>>> lst.insert(0, 9)
>>> lst
______[9, 5, 6, 7, 8, 6]
>>> x = lst.pop(2)
>>> lst
______[9, 5, 7, 8, 6]
>>> lst.remove(x)
>>> lst
______[9, 5, 7, 8]
>>> a, b = lst, lst[:]
>>> a is lst
______True
>>> b == lst
______True
>>> b is lst
______False
Q3: Insert Items
Write a function which takes in a list lst
, an argument entry
, and another argument elem
. This function will check through each item in lst
to see if it is equal to entry
. Upon finding an item equivalent to entry
, the function should modify the list by placing elem
into lst
right after the item. At the end of the function, the modified list should be returned.
See the doctests for examples on how this function is utilized. Use list mutation to modify the original list, no new lists should be created or returned.
Be careful in situations where the values passed into entry
and elem
are equivalent, so as not to create an infinitely long list while iterating through it. If you find that your code is taking more than a few seconds to run, it is most likely that the function is in a loop of inserting new values.
def insert_items(lst, entry, elem):
"""Inserts elem into lst after each occurence of entry and then returns lst.
>>> test_lst = [1, 5, 8, 5, 2, 3]
>>> new_lst = insert_items(test_lst, 5, 7)
>>> new_lst
[1, 5, 7, 8, 5, 7, 2, 3]
>>> large_lst = [1, 4, 8]
>>> large_lst2 = insert_items(large_lst, 4, 4)
>>> large_lst2
[1, 4, 4, 8]
>>> large_lst3 = insert_items(large_lst2, 4, 6)
>>> large_lst3
[1, 4, 6, 4, 6, 8]
>>> large_lst3 is large_lst
True
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q insert_items
Iterators and Generators
Generators also allow us to represent infinite sequences, such as the sequence of natural numbers (1, 2, ...) shown in the function below!
Relevant Topics: Iterators and Generators
def naturals():
"""A generator function that yields the infinite sequence of natural
numbers, starting at 1.
>>> m = naturals()
>>> type(m)
<class 'generator'>
>>> [next(m) for _ in range(10)]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
"""
i = 1
while True:
yield i
i += 1
Q4: Scale
Write a generator function scale(it, multiplier)
which yields the elements of the
iterable it
, multiplied by multiplier
.
As an extra challenge, try writing this function using a yield from
statement!
A yield from
statement yields the values from an iterator one at a time.
def scale(it, multiplier):
"""Yield elements of the iterable it multiplied by a number multiplier.
>>> m = scale([1, 5, 2], 5)
>>> type(m)
<class 'generator'>
>>> list(m)
[5, 25, 10]
>>> m = scale(naturals(), 2)
>>> [next(m) for _ in range(5)]
[2, 4, 6, 8, 10]
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q scale
Q5: Hailstone
Write a generator function that outputs the hailstone sequence starting at number n
.
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.
Note: It is highly encouraged (though not required) to try writing a solution using recursion for some extra practice. Since
hailstone
returns a generator, you can yield from
a call to hailstone
!
def hailstone(n):
"""Yields the elements of the hailstone sequence starting at n.
>>> for num in hailstone(10):
... print(num)
...
10
5
16
8
4
2
1
"""
"*** YOUR CODE HERE ***"
Use Ok to test your code:
python3 ok -q hailstone
Submit
Please fill out the following survey about lab orientation: Survey
Make sure to submit this assignment by running:
python3 ok --submit