Please download lab materials
lab05.zipfrom our QQ group if you don't have one.
In this lab, you have one task:
lab05.py, which is distributed as part of the lab materials in the code directory.Submission: As instructed above, you just need to submit your answer for problems described in section 3 to our OJ website. You may submit more than once before the deadline; only the final submission will be scored. See lab00 for more instructions on submitting assignments.
Readings: You might find the following reference to the textbook useful:
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.
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
nonlocalcannot be used with global variables (names defined in the global frame). If no nonlocal variable is found with the given name, aSyntaxErroris 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.
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:
iter function is called on the iterable to create a corresponding iterator.next function is called on this iterator.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 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
iteron a book gives you a new bookmark independent of other bookmarks, but callingiteron a bookmark gives you the bookmark itself, without changing its position at all. Callingnexton the bookmark moves it to the next page, but does not change the pages in the book. Callingnexton the book wouldn't make sense semantically. We can also have multiple bookmarks, all independent of each other.
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 iterator over f(x) for each x in iterablefilter(f, iterable) - Creates iterator over x for each x in iterable if f(x)zip(iter1, iter2) - Creates iterator over co-indexed pairs (x, y) from both input iterablesreversed(iterable) - Creates iterator over all the elements in the input iterable in - reverse orderlist(iterable) - Creates a list containing all the elements in the input iterabletuple(iterable) - Creates a tuple containing all the elements in the input iterablesorted(iterable) - Creates a sorted list containing all the elements in the input iterableWe 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:
yield statement and returns a generator object.iter function on a generator object returns the same object without modifying its current state.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.next call. Therefore,
next call works like this:
yield.yield statement, but remember the state of the function for future next calls.next calls work like this:
yield statement that was previously executed, and run until the next yield statement.yield statement, but remember the state of the function for future next calls.iter on an iterable object).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
In this section, you are required to complete the problems below and submit your code to Contest lab05 in our OJ website as instructed in lab00 to get your answer scored.
Write a function which takes in an integer n and returns a one-argument function. This function should take in some value x and return n + x the first time it is called, similar to make_adder. The second time it is called, however, it should return n + x + 1, then n + x + 2 the third time, and so on.
def make_adder_inc(n):
"""
>>> adder1 = make_adder_inc(5)
>>> adder2 = make_adder_inc(6)
>>> adder1(2)
7
>>> adder1(2) # 5 + 2 + 1
8
>>> adder1(10) # 5 + 10 + 2
17
>>> [adder1(x) for x in [1, 2, 3]]
[9, 11, 13]
>>> adder2(5)
11
"""
"*** YOUR CODE HERE ***"
Remember to use doctest to test your code:
$ python -m doctest lab05.py
Write a function make_fib that returns a function that returns the next Fibonacci number each time it is called. (The Fibonacci sequence begins with 0 and then 1, after which each element is the sum of the preceding two.) Use a nonlocal statement!
def make_fib():
"""Returns a function that returns the next Fibonacci number
every time it is called.
>>> fib = make_fib()
>>> fib()
0
>>> fib()
1
>>> fib()
1
>>> fib()
2
>>> fib()
3
>>> fib2 = make_fib()
>>> fib() + sum([fib2() for _ in range(5)])
12
>>> from construct_check import check
>>> # Do not use lists in your implementation
>>> check(this_file, 'make_fib', ['List'])
True
"""
"*** YOUR CODE HERE ***"
Generators also allow us to represent infinite sequences, such as the sequence of natural numbers (1, 2, ...).
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
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([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 ***"
Write a generator that outputs the hailstone sequence from hw01.
Here's a quick reminder of how the hailstone sequence is defined:
n as the start.n is even, divide it by 2.n is odd, multiply it by 3 and add 1.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):
"""
>>> for num in hailstone(10):
... print(num)
...
10
5
16
8
4
2
1
"""
"*** YOUR CODE HERE ***"
Note: The code in this question is constructed deliberately in a disgusting way as to cement you knowledge about nonlocal environment diagram. It is not a good coding practice and such kinds of meaningless questions will not appear in our mid-term exam. However, it is a good chance to train your brain to interpret code just as Python.
Draw the environment diagram that results from running the following code.
def moon(f):
sun = 0
moon = [sun]
def run(x):
nonlocal sun, moon
def sun(sun):
return [sun]
y = f(x)
moon.append(sun(y))
return moon[0] and moon[1]
return run
moon(lambda x: moon)(1)
After you've done it on your own, generate an environment diagram in python tutor to check your answer.