Discussion 6: Mutability, Iterators and Generators

This is an online worksheet that you can work on during discussions. Your work is not graded and you do not need to submit anything.

Walkthrough Videos

Feel free to try these problems on the worksheet in discussion or on your own, and then come back to reference these walkthrough videos as you study.

To see these videos, you should be logged into your berkeley.edu email.

YouTube link


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 mutate 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(el): Add el to the end of the list. Return None.
  • extend(lst): Extend the list by concatenating it with lst. Return None.
  • insert(i, el): Insert el at index i. This does not replace any existing elements, but only adds the new element el. Return None.
  • remove(el): Remove the first occurrence of el in list. Errors if el is not in the list. Return None otherwise.
  • pop(i): Remove and return the element at index i.

We can also use list indexing with an assignment statement to change an existing element in a list. For example:

>>> pizza[1] = 'tomatoes'
>>> pizza
['cheese', 'tomatoes', 'onions']

Q1: WWPD: Mutability

What would Python display? In addition to giving the output, draw the box and pointer diagrams for each list to the right.

>>> s1 = [1, 2, 3]
>>> s2 = s1
>>> s1 is s2
>>> s2.extend([5, 6])
>>> s1[4]
>>> s1.append([-1, 0, 1])
>>> s2[5]
[-1, 0, 1]
>>> s3 = s2[:]
>>> s3.insert(3, s2.pop(3))
>>> len(s1)
>>> s1[4] is s3[6]
>>> s3[s2[4][1]]
>>> s1[:3] is s2[:3]
>>> s1[:3] == s2[:3]
>>> s1[4].append(2)
>>> s3[6][3]

Q2: Add This Many

Write a function that takes in a valuex, a value el, and a list s, and adds el to the end of s the number of times x occurs in s. Make sure to modify the original list using list mutation techniques.

Your Answer
Run in 61A Code
def add_this_many(x, el, s):
    """ Adds el to the end of s the number of times x occurs in s.
    >>> s = [1, 2, 4, 2, 1]
    >>> add_this_many(1, 5, s)
    >>> s
    [1, 2, 4, 2, 1, 5, 5]
    >>> add_this_many(2, 2, s)
    >>> s
    [1, 2, 4, 2, 1, 5, 5, 2, 2]
    count = 0
    for element in s:
        if element == x:
            count += 1
    while count > 0:
        count -= 1
Two alternate solutions involve iterating over the list indices and iterating over a copy of the list:
def add_this_many_alt1(x, el, s):
    for i in range(len(s)):
        if s[i] == x:
def add_this_many_alt2(x, el, s):
    for element in list(s):
        if element == x:


An iterable is an object where we can go through its elements one at a time. Specifically, we define an iterable as any object where calling the built-in iter function on it returns an iterator. An iterator is another type of object which can iterate over an iterable by keeping track of which element is next in the iterable.

For example, a sequence of numbers is an iterable, since iter gives us an iterator over the given sequence:

>>> lst = [1, 2, 3]
>>> lst_iter = iter(lst)
>>> lst_iter
<list_iterator object ...>

With an iterator, we can call next on it to get the next element in the iterator. If calling next on an iterator raises a StopIteration exception, this signals to us that the iterator has no more elements to go through. This will be explored in the example below.

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.

For example, we can see what happens when we use iter and next with a list:

>>> lst = [1, 2, 3]
>>> next(lst)             # Calling next on an iterable
TypeError: 'list' object is not an iterator
>>> list_iter = iter(lst) # Creates an iterator for the list
>>> next(list_iter)       # Calling next on an iterator
>>> next(iter(list_iter)) # Calling iter on an iterator returns itself
>>> for e in list_iter:   # Exhausts remainder of list_iter
...     print(e)
>>> next(list_iter)       # No elements left!
>>> lst                   # Original iterable is unaffected
[1, 2, 3]

Q3: WWPD: Iterators

What would Python display?

>>> s = [[1, 2]]
>>> i = iter(s)
>>> j = iter(next(i))
>>> next(j)
>>> s.append(3)
>>> next(i)
>>> next(j)
>>> next(i)


We can define custom iterators by writing a generator function, which returns a special type of iterator called a generator.

A generator function has at least one yield statement and returns a generator object when we call it, without evaluating the body of the generator function itself.

When we first call next on the returned generator, then we will begin evaluating the body of the generator function until an element is yielded or the function otherwise stops (such as if we return). The generator remembers where we stopped, and will continue evaluating from that stopping point on the next time we call next.

As with other iterators, if there are no more elements to be generated, then calling next on the generator will give us a StopIteration.

For example, here's a generator function:

def countdown(n):
    print("Beginning countdown!")
    while n >= 0:
        yield n
        n -= 1

To create a new generator object, we can call the generator function. Each returned generator object from a function call will separately keep track of where it is in terms of evaluating the body of the function. Notice that calling iter on a generator object doesn't create a new bookmark, but simply returns the existing generator object!

>>> c1, c2 = countdown(2), countdown(2)
>>> c1 is iter(c1)  # a generator is an iterator
>>> c1 is c2
>>> next(c1)
Beginning countdown!
>>> next(c2)
Beginning countdown!

In a generator function, we can also have a yield from statement, which will yield each element from an iterator or iterable.

>>> def gen_list(lst):
...     yield from lst
>>> g = gen_list([1, 2])
>>> next(g)
>>> next(g)
>>> next(g)

Q4: Filter-Iter

Implement a generator function called filter_iter(iterable, fn) that only yields elements of iterable for which fn returns True.

Your Answer
Run in 61A Code
def filter_iter(iterable, fn):
    >>> is_even = lambda x: x % 2 == 0
    >>> list(filter_iter(range(5), is_even)) # a list of the values yielded from the call to filter_iter
    [0, 2, 4]
    >>> all_odd = (2*y-1 for y in range(5))
    >>> list(filter_iter(all_odd, is_even))
    >>> naturals = (n for n in range(1, 100))
    >>> s = filter_iter(naturals, is_even)
    >>> next(s)
    >>> next(s)
    for elem in iterable:
        if fn(elem):
            yield elem

Q5: Merge

Write a generator function merge that takes in two infinite generators a and b that are in increasing order without duplicates and returns a generator that has all the elements of both generators, in increasing order, without duplicates.

Your Answer
Run in 61A Code
def merge(a, b):
    >>> def sequence(start, step):
    ...     while True:
    ...         yield start
    ...         start += step
    >>> a = sequence(2, 3) # 2, 5, 8, 11, 14, ...
    >>> b = sequence(3, 2) # 3, 5, 7, 9, 11, 13, 15, ...
    >>> result = merge(a, b) # 2, 3, 5, 7, 8, 9, 11, 13, 14, 15
    >>> [next(result) for _ in range(10)]
    [2, 3, 5, 7, 8, 9, 11, 13, 14, 15]
    first_a, first_b = next(a), next(b)
    while True:
        if first_a == first_b:
            yield first_a
            first_a, first_b = next(a), next(b)
        elif first_a < first_b:
            yield first_a
            first_a = next(a)
            yield first_b
            first_b = next(b)

Q6: Primes Generator

Write a function primes_gen that takes a single argument n and yields all prime numbers less than or equal to n in decreasing order. Assume n >= 1. You may use the is_prime function included below, which we implemented in Discussion 3.

Optional Challenge: Now rewrite the generator so that it also prints the primes in ascending order.

Your Answer
Run in 61A Code
def is_prime(n):
    """Returns True if n is a prime number and False otherwise.
    >>> is_prime(2)
    >>> is_prime(16)
    >>> is_prime(521)
    def helper(i):
        if i > (n ** 0.5): # Could replace with i == n
            return True
        elif n % i == 0:
            return False
        return helper(i + 1)
    return helper(2)

def primes_gen(n):
    """Generates primes in decreasing order.
    >>> pg = primes_gen(7)
    >>> list(pg)
    [7, 5, 3, 2]
    if n == 1:
    if is_prime(n):
        yield n
    yield from primes_gen(n-1)

Q7: (Optional) Mystery Reverse Environment Diagram

Fill in the lines below so that the variables in the global frame are bound to the values below. Note that the image does not contain a full environment diagram. You may only use brackets, colons, p and q in your answer.

Hint: If you get stuck, feel free to try out different combinations in PythonTutor!


Your Answer
Run in 61A Code
def mystery(p, q):

p = [2, 3]
q = [4, [p]]
mystery(q, p)