问题1:坐标轴及标签显示
正如你所看到的,他最近运气不佳。他想用一些精选的表情符号沿着在推特上发布这条消息,但是,就像现在看起来的那样,他的追随者可能会觉得很困惑。他问你是否可以帮助他做出以下改变:
添加标题“500次老虎机抽奖的结果”
使y轴从0开始。
将标签“Balance”添加到y轴
在调用type(graph)之后,您可以看到Jimmy的图是类型 matplotlib.axes._subplots.AxesSubplot 。嗯,这是新的。通过调用dir(graph),您会发现三个看起来很有用的方法:.set_title()、.set_ylim()和.set_ylabel()。
使用这些方法根据Jimmy的请求完成函数prettify_graph。我们已经为您检查了第一个请求(设置标题)。
(记住:如果你不知道这些方法的作用,请使用help()函数!)
.set_title()
:设置标题.set_ylim()
:设置y轴.set_ylabel()
:设置y轴标签
def prettify_graph(graph):
"""Modify the given graph according to Jimmy's requests: add a title, make the y-axis
start at 0, label the y-axis. (And, if you're feeling ambitious, format the tick marks
as dollar amounts using the "$" symbol.)
"""
graph.set_title("Results of 500 slot machine pulls")
# Complete steps 2 and 3 here
# graph.set_title("Results of 500 slot machine pulls")
graph.set_ylim(0)
graph.set_ylabel("Balance")
graph = jimmy_slots.get_graph()
prettify_graph(graph)
graph
额外好处:您可以对y轴上的数字进行格式设置,使它们看起来像美元金额吗?例如,200美元而不是200美元。
def prettify_graph(graph):
graph.set_title("Results of 500 slot machine pulls")
# Make the y-axis begin at 0
graph.set_ylim(bottom=0)
# Label the y-axis
graph.set_ylabel("Balance")
# Bonus: format the numbers on the y-axis as dollar amounts
# An array of the values displayed on the y-axis (150, 175, 200, etc.)
ticks = graph.get_yticks()
# Format those values into strings beginning with dollar sign
new_labels = ['${}'.format(int(amt)) for amt in ticks]
# Set the new labels
graph.set_yticklabels(new_labels)
graph = jimmy_slots.get_graph()
prettify_graph(graph)
graph
问题2:赢得比赛的最佳项目并计数
Luigi正试图进行分析,以确定在Mario Kart赛道上赢得比赛的最佳项目。他有一些字典列表形式的数据,看起来像…
[
{‘name’: ‘Peach’, ‘items’: [‘green shell’, ‘banana’, ‘green shell’,], ‘finish’: 3},
{‘name’: ‘Bowser’, ‘items’: [‘green shell’,], ‘finish’: 1},
# Sometimes the racer’s name wasn’t recorded
{‘name’: None, ‘items’: [‘mushroom’,], ‘finish’: 2},
{‘name’: ‘Toad’, ‘items’: [‘green shell’, ‘mushroom’], ‘finish’: 1},
]
“items”是参赛者在比赛中获得的所有能量提升物品的列表,“finish”是他们在比赛中的位置(1代表第一名,3代表第三名,等等)。
他写了下面的函数来获取这样的列表,并返回一个字典,将每个条目映射到第一名完成者拾取的次数。
def best_items(racers):
# 创建空字典用于存储第一名完成者拾取的次数
winner_item_counts = {}
for i in range(len(racers)):
# 对于列表中每个字典进行操作
racer = racers[i]
# 如果是第一名
if racer['finish'] == 1:
# 获取第一名的项目
for item in racer['items']:
# 如果结果字典中没有该元素,进行创建
if item not in winner_item_counts:
winner_item_counts[item] = 0
# 有该元素则次数加1
winner_item_counts[item] += 1
# 如果名字未记录,发出警告
if racer['name'] is None:
print("警告:{}/{}名字为空,未记录! (racer = {})".format((i+1),len(racers),racer['name']))
return winner_item_counts
sample = [
{'name': 'Peach', 'items': ['green shell', 'banana', 'green shell',], 'finish': 3},
{'name': 'Bowser', 'items': ['green shell',], 'finish': 1},
{'name': None, 'items': ['mushroom',], 'finish': 2},
{'name': 'Toad', 'items': ['green shell', 'mushroom'], 'finish': 1},
]
best_items(sample)
问题3:21点游戏
假设我们想创建一个新类型来表示21点中的手牌。我们可能想对这种类型做的一件事是重载比较运算符,如>和<=,以便我们可以使用它们来检查一只手是否击败另一只手。如果我们能做到这一点,那就太酷了:
hand1 = BlackjackHand([‘K’, ‘A’])
hand2 = BlackjackHand([‘7’, ‘10’, ‘A’])
hand1 > hand2
True
好吧,我们不会在这个问题中完成所有这些(定义自定义类有点超出了这些课程的范围),但是我们要求您在下面的函数中编写的代码与我们定义自己的BlackjackHand类时必须编写的代码非常相似。(We我把它放在__gt__magic方法中,为>定义我们的自定义行为。)
根据文档字符串填充blackjack_hand_greater_than函数体
方法1:把超过21点的最后记为0
1、首先定义一个21点的计数函数BlackjackHand(hand)
def BlackjackHand(hand):
count = 0
A_count = 0
# 进行遍历,如果为A,J,Q,K中的其中一个,加各自对应的数
# 若为1~10,则加int(x)
for x in hand:
if x == 'A':
A_count = A_count + 1
count = count + 11
continue
if (x == 'J' or x == 'Q' or x == 'K'):
count = count + 10
continue
else:
count = count + int(x)
# 判断
if A_count == 0 and count <= 21:
count = count
elif A_count == 0 and count > 21:
count = 0
elif A_count > 0 and count <= 21:
count = count
elif A_count > 0 and count > 21:
for i in range(A_count):
count = count - 10
if count <= 21:
count = count
if count > 21:
count = 0
return count
# 测试
print(BlackjackHand(['K']))
print(BlackjackHand(['3', '4']))
print(BlackjackHand(['10']))
print(BlackjackHand(['K', 'K', '2']))
print(BlackjackHand(['A', 'A', '9']))
print(BlackjackHand(['A', 'A', '9', '3']))
print(BlackjackHand(['9','Q',' 8','A']))
2、最终代码
def blackjack_hand_greater_than(hand_1, hand_2):
"""
Return True if hand_1 beats hand_2, and False otherwise.
In order for hand_1 to beat hand_2 the following must be true:
- The total of hand_1 must not exceed 21
- The total of hand_1 must exceed the total of hand_2 OR hand_2's total must exceed 21
Hands are represented as a list of cards. Each card is represented by a string.
When adding up a hand's total, cards with numbers count for that many points. Face
cards ('J', 'Q', and 'K') are worth 10 points. 'A' can count for 1 or 11.
When determining a hand's total, you should try to count aces in the way that
maximizes the hand's total without going over 21. e.g. the total of ['A', 'A', '9'] is 21,
the total of ['A', 'A', '9', '3'] is 14.
Examples:
>>> blackjack_hand_greater_than(['K'], ['3', '4'])
True
>>> blackjack_hand_greater_than(['K'], ['10'])
False
>>> blackjack_hand_greater_than(['K', 'K', '2'], ['3'])
False
"""
if BlackjackHand(hand_1) > BlackjackHand(hand_2):
return True
else:
return False
def BlackjackHand(hand):
count = 0
A_count = 0
for x in hand:
if x == 'A':
A_count = A_count + 1
count = count + 11
continue
if (x == 'J' or x == 'Q' or x == 'K'):
count = count + 10
continue
else:
count = count + int(x)
if A_count == 0 and count <= 21:
count = count
elif A_count == 0 and count > 21:
count = 0
elif A_count > 0 and count <= 21:
count = count
elif A_count > 0 and count > 21:
for i in range(A_count):
count = count - 10
if count <= 21:
count = count
if count > 21:
count = 0
return count
方法2:把超过21点在最后进行判断
1、首先定义一个21点的计数函数BlackjackHand(hand)
def BlackjackHand(hand):
count = 0
A_count = 0
for x in hand:
if x == 'A':
A_count = A_count + 1
count = count + 11
continue
if (x == 'J' or x == 'Q' or x == 'K'):
count = count + 10
continue
else:
count = count + int(x)
if A_count > 0 and count > 21:
for i in range(A_count):
count = count - 10
if count <= 21:
count = count
if count > 21:
count = count
return count
# 测试
print(BlackjackHand(['K']))
print(BlackjackHand(['3', '4']))
print(BlackjackHand(['10']))
print(BlackjackHand(['K', 'K', '2']))
print(BlackjackHand(['A', 'A', '9']))
print(BlackjackHand(['A', 'A', '9', '3']))
print(BlackjackHand(['9','Q',' 8','A']))
超过21点不变,等到最后进行判断为False
2、最终代码
def blackjack_hand_greater_than(hand_1, hand_2):
"""
如果hand_1打败hand_2返回True,否则返回False。
为了使hand_1打败hand_2,以下条件必须成立:
—hand_1的总数不能超过21
—hand_1的总和必须大于hand_2的总和,或者hand_2的总和必须大于21
手牌被表示为一张牌的列表。每张卡片由一个字符串表示。
当把一手牌的总牌数加起来时,带有数字的牌就会得到相应的点数。脸
卡片(“J”、“Q”和“K”)值10分。“A”可以数1或11。
当决定一手牌的总数时,你应该尝试用下面的方法来计算a
在不超过21的情况下最大化手牌总数。例如,['A', 'A', '9']的总数是21,
['A', 'A', '9', '3']的总数是14。
Examples:
>>> blackjack_hand_greater_than(['K'], ['3', '4'])
True
>>> blackjack_hand_greater_than(['K'], ['10'])
False
>>> blackjack_hand_greater_than(['K', 'K', '2'], ['3'])
False
"""
if (BlackjackHand(hand_1)<=21) and (BlackjackHand(hand_1) > BlackjackHand(hand_2)) or (BlackjackHand(hand_1)<=21 and BlackjackHand(hand_2)>21):
return True
else:
return False
def BlackjackHand(hand):
# 定义总数和'A'的数量
count = 0
A_count = 0
#
for x in hand:
if x == 'A':
A_count = A_count + 1
count = count + 11
continue
if (x == 'J' or x == 'Q' or x == 'K'):
count = count + 10
continue
else:
count = count + int(x)
# 判断
if A_count > 0 and count > 21:
for i in range(A_count):
count = count - 10
if count <= 21:
count = count
if count > 21:
count = count
return count
方法3:官方代码与方法2相似
def hand_total(hand):
"""Helper function to calculate the total points of a blackjack hand.
"""
total = 0
# 计算a的数量,并在最后处理如何使用它们
aces = 0
for card in hand:
if card in ['J', 'Q', 'K']:
total += 10
elif card == 'A':
aces += 1
else:
# Convert number cards (e.g. '7') to ints
total += int(card)
# At this point, total is the sum of this hand's cards *not counting aces*.
# 添加a,现在把它们记为1。这是我们从这只手能得到的最小的总数
total += aces
# 把a从1 “升级”到11只要它能帮我们接近21
# without busting
while total + 10 <= 21 and aces > 0:
# Upgrade an ace from 1 to 11
total += 10
aces -= 1
return total
def blackjack_hand_greater_than(hand_1, hand_2):
total_1 = hand_total(hand_1)
total_2 = hand_total(hand_2)
return total_1 <= 21 and (total_1 > total_2 or total_2 > 21)
简化:
if card in ['J', 'Q', 'K']: