[2021 spring] CS61A Lab 5: Python Lists, Data Abstraction, Trees
2021/6/29 9:52:47
本文主要是介绍[2021 spring] CS61A Lab 5: Python Lists, Data Abstraction, Trees,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
lab05: https://inst.eecs.berkeley.edu/~cs61a/sp21/lab/lab05/#topics
lab5包括对列表的理解,数据抽象,和树
- Topics
- List Comprehensions
- Data Abstraction
- Trees
- Required Questions
- Q1: Couple(List Comprehensions)
- Q2: Distance(Data Abstraction)
- Q3: Closer city(Data Abstraction)
- Q5: Finding Berries!(Trees)
- Q6: Sprout leaves(Trees)
Topics
List Comprehensions
Data Abstraction
Trees
Required Questions
Q1: Couple(List Comprehensions)
def couple(s, t): """Return a list of two-element lists in which the i-th element is [s[i], t[i]]. >>> a = [1, 2, 3] >>> b = [4, 5, 6] >>> couple(a, b) [[1, 4], [2, 5], [3, 6]] >>> c = ['c', 6] >>> d = ['s', '1'] >>> couple(c, d) [['c', 's'], [6, '1']] """ assert len(s) == len(t) "*** YOUR CODE HERE ***" return [[s[i], t[i]] for i in range(len(s))]
Q2: Distance(Data Abstraction)
from math import sqrt def distance(city_a, city_b): """ >>> city_a = make_city('city_a', 0, 1) >>> city_b = make_city('city_b', 0, 2) >>> distance(city_a, city_b) 1.0 >>> city_c = make_city('city_c', 6.5, 12) >>> city_d = make_city('city_d', 2.5, 15) >>> distance(city_c, city_d) 5.0 """ "*** YOUR CODE HERE ***" return sqrt((get_lat(city_a) - get_lat(city_b)) ** 2 +\ (get_lon(city_a) - get_lon(city_b)) ** 2)
Q3: Closer city(Data Abstraction)
def closer_city(lat, lon, city_a, city_b): """ Returns the name of either city_a or city_b, whichever is closest to coordinate (lat, lon). If the two cities are the same distance away from the coordinate, consider city_b to be the closer city. >>> 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 ***" target = make_city('target', lat, lon) dist_a = distance(target, city_a) dist_b = distance(target, city_b) return get_name(city_a) if dist_a <= dist_b else get_name(city_b)
Q5: Finding Berries!(Trees)
def berry_finder(t): """Returns True if t contains a node with the value 'berry' and False otherwise. >>> scrat = tree('berry') >>> berry_finder(scrat) True >>> sproul = tree('roots', [tree('branch1', [tree('leaf'), tree('berry')]), tree('branch2')]) >>> berry_finder(sproul) True >>> numbers = tree(1, [tree(2), tree(3, [tree(4), tree(5)]), tree(6, [tree(7)])]) >>> berry_finder(numbers) False >>> t = tree(1, [tree('berry',[tree('not berry')])]) >>> berry_finder(t) True """ "*** YOUR CODE HERE ***" if label(t) == 'berry': return True for c in branches(t): if berry_finder(c): return True return False
Q6: Sprout leaves(Trees)
def sprout_leaves(t, leaves): """Sprout new leaves containing the data in leaves at each leaf in the original tree t and return the resulting tree. >>> t1 = tree(1, [tree(2), tree(3)]) >>> print_tree(t1) 1 2 3 >>> new1 = sprout_leaves(t1, [4, 5]) >>> print_tree(new1) 1 2 4 5 3 4 5 >>> t2 = tree(1, [tree(2, [tree(3)])]) >>> print_tree(t2) 1 2 3 >>> new2 = sprout_leaves(t2, [6, 1, 2]) >>> print_tree(new2) 1 2 3 6 1 2 """ "*** YOUR CODE HERE ***" if is_leaf(t): return tree(label(t), [tree(i) for i in leaves]) return tree(label(t), [sprout_leaves(c, leaves) for c in branches(t)])
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