Lab 4 - Lists practice, Data abstraction

Quite an easy lab, if you've ever used a list/array...

Instructions: https://inst.eecs.berkeley.edu/~cs61a/su19/lab/lab04/

Solution: https://github.com/tomthestrom/cs61a/blob/master/lab/lab04/lab04.py

Q1 is no coding, deals with retrieving vals by index - WWPD (What Would Python Display?)

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.

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 ***"

Q2: Solution

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 [[x, y] for x, y in zip(s1, s2)]

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]]

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 ***"

Q3: Solution

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 ***"
    enumerated = []
    for i in range(len(s)):
        enumerated.append([start + i, s[i]])
    return enumerated

City Data Abstraction

More here: https://inst.eecs.berkeley.edu/~cs61a/su19/lab/lab04/#city-data-abstraction

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!

Q4: Distance

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 ***"

Q4: Solution

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 ***"
    sub_pow_coord = lambda coord1, coord2: pow((coord1 - coord2), 2)
    
    return sqrt(sub_pow_coord(
        get_lat(city1),
        get_lat(city2))
                +
                sub_pow_coord(
        get_lon(city1),
        get_lon(city2))
               )

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.

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 ***"

Q5: Solution

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 ***"
    coords = make_city('coords', lat, lon)
        
    return get_name(city1) if distance(city1, coords) < distance(city2, coords) else get_name(city2)

Q6: Don't violate the abstraction barrier!

No Code - just a test that checks whether you've written the previous 2 correctly - using constructors, selectors, etc - manipulating data as a compound value.