标签：city Abstraction return Python leaves tree Lab berry make
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)
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)])
标签：city,Abstraction,return,Python,leaves,tree,Lab,berry,make 来源： https://www.cnblogs.com/ikventure/p/14940221.html
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