网站开发后台php技术专业的网络推广

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-axisstart at 0, label the y-axis. (And, if you're feeling ambitious, format the tick marksas 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 0graph.set_ylim(bottom=0)# Label the y-axisgraph.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 signnew_labels = ['${}'.format(int(amt)) for amt in ticks]# Set the new labelsgraph.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1winner_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 = 0A_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 + 1count = count + 11continueif (x == 'J' or  x == 'Q' or x == 'K'):count = count + 10continueelse:count = count + int(x)# 判断        if A_count == 0 and count <= 21:count = countelif A_count == 0 and count > 21:count = 0elif A_count > 0 and count <= 21:count = countelif A_count > 0 and count > 21:for i in range(A_count):count = count - 10if count <= 21:count = countif count > 21:count = 0return 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 21Hands 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. Facecards ('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 Trueelse:return Falsedef BlackjackHand(hand):count = 0A_count = 0for x in hand:if x == 'A':A_count = A_count + 1count = count + 11continueif (x == 'J' or  x == 'Q' or x == 'K'):count = count + 10continueelse:count = count + int(x)if A_count == 0 and count <= 21:count = countelif A_count == 0 and count > 21:count = 0elif A_count > 0 and count <= 21:count = countelif A_count > 0 and count > 21:for i in range(A_count):count = count - 10if count <= 21:count = countif count > 21:count = 0return 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 = 0A_count = 0for x in hand:if x == 'A':A_count = A_count + 1count = count + 11continueif (x == 'J' or  x == 'Q' or x == 'K'):count = count + 10continueelse:count = count + int(x)if A_count > 0 and count > 21:for i in range(A_count):count = count - 10if count <= 21:count = countif count > 21:count = countreturn 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 Trueelse:return Falsedef BlackjackHand(hand):# 定义总数和'A'的数量count = 0A_count = 0# for x in hand:if x == 'A':A_count = A_count + 1count = count + 11continueif (x == 'J' or  x == 'Q' or x == 'K'):count = count + 10continueelse:count = count + int(x)# 判断        if A_count > 0 and count > 21:for i in range(A_count):count = count - 10if count <= 21:count = countif count > 21:count = countreturn 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 = 0for card in hand:if card in ['J', 'Q', 'K']:total += 10elif card == 'A':aces += 1else:# Convert number cards (e.g. '7') to intstotal += 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 bustingwhile total + 10 <= 21 and aces > 0:# Upgrade an ace from 1 to 11total += 10aces -= 1return totaldef 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']:
•