Lab 4: Lists, and Data Abstraction

Due at 11:59pm on Friday, 07/12/2019.

Starter Files

Download lab04.zip. Inside the archive, you will find starter files for the questions in this lab, along with a copy of the Ok autograder.

Submission

By the end of this lab, you should have submitted the lab with python3 ok --submit. You may submit more than once before the deadline; only the final submission will be graded. Check that you have successfully submitted your code on okpy.org.

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.

Lists

Lists are Python data structures that can store multiple values. Each value can be any type and can even be another list! A list is written as a comma separated list of expressions within square brackets:

>>> list_of_nums = [1, 2, 3, 4]
>>> list_of_bools = [True, True, False, False]
>>> nested_lists = [1, [2, 3], [4, [5]]]

Each element in a list is assigned an index. Lists are zero-indexed, meaning their indices start at 0 and increase in sequential order. To retrieve an element from a list, use list indexing:

>>> lst = [6, 5, 4, 3, 2, 1]
>>> lst[0]
6
>>> lst[3]
3

Often times we need to know how long a list is when we're working with it. To find the length of a list, call the function len on it:

>>> len([])
0
>>> len([2, 4, 6, 8, 10])
5

Tip: Recall that empty lists, [], are false-y values. Therefore, you can use an if statement like the following if you only want to do operations on non-empty lists:

if lst:
    # Do stuff with the elements of list

This is equivalent to:

if len(lst) > 0:
    # Do stuff

You can also create a copy of some portion of the list using list slicing. To slice a list, use this syntax: lst[<start index>:<end index>]. This expression evaluates to a new list containing the elements of lst starting at and including the element at <start index> up to but not including the element at end index.

>>> lst = [True, False, True, True, False]
>>> lst[1:4]
[False, True, True]
>>> lst[:3]  # Start index defaults to 0
[True, False, True]
>>> lst[3:]  # End index defaults to len(lst)
[True, False]
>>> lst[:]  # Creates a copy of the whole list
[True, False, True, True, False]

List Comprehensions

List comprehensions are a compact and powerful way of creating new lists out of sequences. The general syntax for a list comprehension is the following:

[<expression> for <element> in <sequence> if <conditional>]

The syntax is designed to read like English: "Compute the expression for each element in the sequence if the conditional is true for that element."

Let's see it in action:

>>> [i**2 for i in [1, 2, 3, 4] if i % 2 == 0]
[4, 16]

Here, for each element i in [1, 2, 3, 4] that satisfies i % 2 == 0, we evaluate the expression i**2 and insert the resulting values into a new list. In other words, this list comprehension will create a new list that contains the square of each of the even elements of the original list.

If we were to write this using a for statement, it would look like this:

>>> lst = []
>>> for i in [1, 2, 3, 4]:
...     if i % 2 == 0:
...         lst += [i**2]
>>> lst
[4, 16]

Note: The if clause in a list comprehension is optional. For example, you can just say:

>>> [i**2 for i in [1, 2, 3, 4]]
[1, 4, 9, 16]

Data Abstraction

Data abstraction is a powerful concept in computer science that allows programmers to treat code as objects -- for example, car objects, chair objects, people objects, etc. That way, programmers don't have to worry about how code is implemented -- they just have to know what it does.

Data abstraction mimics how we think about the world. When you want to drive a car, you don't need to know how the engine was built or what kind of material the tires are made of. You just have to know how to turn the wheel and press the gas pedal.

An abstract data type consists of two types of functions:

  • Constructors: functions that build the abstract data type.
  • Selectors: functions that retrieve information from the data type.

Programmers design ADTs to abstract away how information is stored and calculated such that the end user does not need to know how constructors and selectors are implemented. The nature of abstract data types allows whoever uses them to assume that the functions have been written correctly and work as described.

Required Questions

Lists Practice

Q1: List Indexing

Use Ok to test your knowledge with the following "List Indexing" questions:

python3 ok -q indexing -u

For each of the following lists, what is the list indexing expression that evaluates to 7? For example, if x = [7], then the answer would be x[0]. You can use the interpreter or Python Tutor to experiment with your answers.

>>> x = [1, 3, [5, 7], 9]
______
x[2][1]
>>> x = [[3, [5, 7], 9]]
______
x[0][1][1]

What would Python display? If you get stuck, try it out in the Python interpreter!

>>> lst = [3, 2, 7, [84, 83, 82]]
>>> lst[4]
______
Error
>>> lst[3][0]
______
84

Q2: Couple

Implement the function couple, which takes in two lists and returns a list that contains lists with i-th elements of two sequences coupled together. You can assume the lengths of two sequences are the same. Try using a list comprehension.

Hint: You may find the built in range function helpful.

def couple(s1, s2):
    """Return a list that contains lists with i-th elements of two sequences
    coupled together.
    >>> s1 = [1, 2, 3]
    >>> s2 = [4, 5, 6]
    >>> couple(s1, s2)
    [[1, 4], [2, 5], [3, 6]]
    >>> s3 = ['c', 6]
    >>> s4 = ['s', '1']
    >>> couple(s3, s4)
    [['c', 's'], [6, '1']]
    """
    assert len(s1) == len(s2)
"*** YOUR CODE HERE ***"
return [[s1[i], s2[i]] for i in range(0, len(s1))]

Use Ok to test your code:

python3 ok -q couple

Q3: Enumerate

Implement enumerate, which pairs the elements of a sequence with their indices, offset by a starting value. enumerate takes a sequence s and a starting value start. It returns a list of pairs, in whichthe i-th element is i + start paired with the i-th element of s. For example:

>>> enumerate(['maps', 21, 47], start=1)
>>> [[1, 'maps'], [2, 21], [3, 47]]

Hint: Consider using couple from Question 2!

Hint 2: You may find the built in range function helpful.

def enumerate(s, start=0):
    """Returns a list of lists, where the i-th list contains i+start and
    the i-th element of s.
    >>> enumerate([6, 1, 'a'])
    [[0, 6], [1, 1], [2, 'a']]
    >>> enumerate('five', 5)
    [[5, 'f'], [6, 'i'], [7, 'v'], [8, 'e']]
    """
"*** YOUR CODE HERE ***"
return couple(range(start, start+len(s)), s)

Use Ok to test your code:

python3 ok -q enumerate

City Data Abstraction

Say we have an abstract data type for cities. A city has a name, a latitude coordinate, and a longitude coordinate.

Our ADT has one constructor:

  • make_city(name, lat, lon): Creates a city object with the given name, latitude, and longitude.

We also have the following selectors in order to get the information for each city:

  • get_name(city): Returns the city's name
  • get_lat(city): Returns the city's latitude
  • get_lon(city): Returns the city's longitude

Here is how we would use the constructor and selectors to create cities and extract their information:

>>> berkeley = make_city('Berkeley', 122, 37)
>>> get_name(berkeley)
'Berkeley'
>>> get_lat(berkeley)
122
>>> new_york = make_city('New York City', 74, 40)
>>> get_lon(new_york)
40

All of the selector and constructor functions can be found in the lab file, if you are curious to see how they are implemented. However, the point of data abstraction is that we do not need to know how an abstract data type is implemented, but rather just how we can interact with and use the data type.

Q4: Distance

We will now implement the function distance, which computes the distance between two city objects. Recall that the distance between two coordinate pairs (x1, y1) and (x2, y2) can be found by calculating the sqrt of (x1 - x2)**2 + (y1 - y2)**2. We have already imported sqrt for your convenience. Use the latitude and longitude of a city as its coordinates; you'll need to use the selectors to access this info!

from math import sqrt
def distance(city1, city2):
    """
    >>> city1 = make_city('city1', 0, 1)
    >>> city2 = make_city('city2', 0, 2)
    >>> distance(city1, city2)
    1.0
    >>> city3 = make_city('city3', 6.5, 12)
    >>> city4 = make_city('city4', 2.5, 15)
    >>> distance(city3, city4)
    5.0
    """
"*** YOUR CODE HERE ***"
lat_1, lon_1 = get_lat(city1), get_lon(city1) lat_2, lon_2 = get_lat(city2), get_lon(city2) return sqrt((lat_1 - lat_2)**2 + (lon_1 - lon_2)**2) # Video walkthrough: https://youtu.be/4vB06Ymrico

Use Ok to test your code:

python3 ok -q distance

Q5: Closer city

Next, implement closer_city, a function that takes a latitude, longitude, and two cities, and returns the name of the city that is relatively closer to the provided latitude and longitude.

You may only use the selectors and constructors introduced above and the distance function you just defined for this question.

Hint: How can use your distance function to find the distance between the given location and each of the given cities?

def closer_city(lat, lon, city1, city2):
    """
    Returns the name of either city1 or city2, whichever is closest to
    coordinate (lat, lon).

    >>> berkeley = make_city('Berkeley', 37.87, 112.26)
    >>> stanford = make_city('Stanford', 34.05, 118.25)
    >>> closer_city(38.33, 121.44, berkeley, stanford)
    'Stanford'
    >>> bucharest = make_city('Bucharest', 44.43, 26.10)
    >>> vienna = make_city('Vienna', 48.20, 16.37)
    >>> closer_city(41.29, 174.78, bucharest, vienna)
    'Bucharest'
    """
"*** YOUR CODE HERE ***"
new_city = make_city('arb', lat, lon) dist1 = distance(city1, new_city) dist2 = distance(city2, new_city) if dist1 < dist2: return get_name(city1) return get_name(city2) # Video walkthrough: https://youtu.be/Owwdz_Km87g

Use Ok to test your code:

python3 ok -q closer_city

Q6: Don't violate the abstraction barrier!

Note: this question has no code-writing component (if you implemented distance and closer_city correctly!)

When writing functions that use an ADT, we should use the constructor(s) and selector(s) whenever possible instead of assuming the ADT's implementation. Relying on a data abstraction's underlying implementation is known as violating the abstraction barrier, and we never want to do this!

It's possible that you passed the doctests for distance and closer_city even if you violated the abstraction barrier. To check whether or not you did so, run the following command:

Use Ok to test your code:

python3 ok -q check_abstraction

The make_check_abstraction function exists only for the doctest, which swaps out the implementations of the city abstraction with something else, runs the tests from the previous two parts, then restores the original abstraction.

The nature of the abstraction barrier guarantees that changing the implementation of an ADT shouldn't affect the functionality of any programs that use that ADT, as long as the constructors and selectors were used properly.

If you passed the Ok tests for the previous questions but not this one, the fix is simple! Just replace any code that violates the abstraction barrier, i.e. creating a city with a new list object or indexing into a city, with the appropriate constructor or selector.

Make sure that your functions pass the tests with both the first and the second implementations of the City ADT and that you understand why they should work for both before moving on.

Optional Question

This question can be found in lab04_extra.py.

Q7: Squares only

Implement the function squares, which takes in a list of positive integers. It returns a list that contains the square roots of the elements of the original list that are perfect squares. Try using a list comprehension.

You may find the round function useful.

>>> round(10.5)
10
>>> round(10.51)
11
def squares(s):
    """Returns a new list containing square roots of the elements of the
    original list that are perfect squares.

    >>> seq = [8, 49, 8, 9, 2, 1, 100, 102]
    >>> squares(seq)
    [7, 3, 1, 10]
    >>> seq = [500, 30]
    >>> squares(seq)
    []
    """
"*** YOUR CODE HERE ***"
return [round(n ** 0.5) for n in s if n == round(n ** 0.5) ** 2]

It might be helpful to construct a skeleton list comprehension to begin with:

[sqrt(x) for x in s if is_perfect_square(x)]

This is great, but it requires that we have an is_perfect_square function. How might we check if something is a perfect square?

  • If the square root of a number is a whole number, then it is a perfect square. For example, sqrt(61) = 7.81024... (not a perfect square) and sqrt(49) = 7 (perfect square).
  • Once we obtain the square root of the number, we just need to check if something is a whole number. The is_perfect_square function might look like:

    def is_perfect_square(x):
        return is_whole(sqrt(x))
  • One last piece of the puzzle: to check if a number is whole, we just need to see if it has a decimal or not. The way we've chosen to do it in the solution is to compare the original number to the round version (thus removing all decimals), but a technique employing floor division (//) or something else entirely could work too.

We've written all these helper functions to solve this problem, but they are actually all very short. Therefore, we can just copy the body of each into the original list comprehension, arriving at the solution we finally present.

Video walkthrough: https://youtu.be/YwLFB9paET0

Use Ok to test your code:

python3 ok -q squares

Q8: Key of Min Value

The built-in min function takes a collection of elements (such as a list or a dictionary) and returns the collection's smallest element. The min function also takes in an optional keyword argument called key, which must be a one-argument function. The key function is called on each element of the collection, and the return values are used for comparison. For example:

>>> min([-1, 0, 1])                    # no key argument; return smallest input
-1
>>> min([-1, 0, 1], key=lambda x: x*x) # return input with the smallest square
0

Implement key_of_min_value, which takes in a dictionary d and returns the key that corresponds to the minimum value in d. This behavior differs from just calling min on a dictionary, which would return the smallest key. Make sure your solution is only one line and uses the min function.

def key_of_min_value(d):
    """Returns the key in a dict d that corresponds to the minimum value of d.
    >>> letters = {'a': 6, 'b': 5, 'c': 4, 'd': 5}
    >>> min(letters)
    'a'
    >>> key_of_min_value(letters)
    'c'
    """
"*** YOUR CODE HERE ***"
return min(d, key=lambda x: d[x])

Use Ok to test your code:

python3 ok -q key_of_min_value