if -statements，但是不确定如何更改代码段以减少使用。import random
def playchosengame(chosengame):
if chosengame == 1:
doorchoice = int(raw_input("How many doors would you like to play with? > "))
onegameatatime(doorchoice)
elif chosengame == 2:
numberofgames = int(raw_input("How many games would you like to play? > "))
strategy = int(raw_input("Would you like to always change your answer (1) or keep it the same (2)? > "))
doorchoice = int(raw_input("How many doors would you like to play with? > "))
onestrategy(numberofgames,strategy,doorchoice)
else:
print "What did you say?"
def onestrategy(chosennumberofgames,chosenstrategy,doorchoice):
# for playing x number of Monty Hall games with one type of strategy
wincount = 0
i = 1
for i in range(1,chosennumberofgames + 1):
possibleanswers = range(1,doorchoice + 1)
correctanswer = random.randint(1,doorchoice)
incorrectpossibilities = possibleanswers
incorrectpossibilities.remove(correctanswer)
youranswer = random.randint(1,doorchoice)
if youranswer == correctanswer:
otherremaining = random.choice(incorrectpossibilities)
else:
otherremaining = youranswer
incorrectpossibilities.remove(otherremaining)
print "The correct answer is NOT one of these: %r" % incorrectpossibilities
if youranswer == correctanswer:
print "Which means it is either your answer: %r, or the only other remaining option, %r." % (correctanswer, otherremaining)
if chosenstrategy == 1:
finalanswer = otherremaining
else:
finalanswer = correctanswer
else:
print "Which means it is either your answer: %r, or the only other remaining option, %r." % (youranswer, correctanswer)
if chosenstrategy == 1:
finalanswer = correctanswer
else:
finalanswer = youranswer
print "You chose %r" % finalanswer
if finalanswer == correctanswer:
wincount += 1
print "You win!"
else:
print "You lose!"
i += 1
print "You won %r out of %r games. Congrats!" % (wincount, i - 1)
def onegameatatime(doorchoice):
# for playing one Monty Hall game at a time.
playagain = 1
while playagain == 1:
possibleanswers = range(1,doorchoice + 1)
correctanswer = random.randint(1,doorchoice)
incorrectpossibilities = possibleanswers
incorrectpossibilities.remove(correctanswer)
youranswer = int(raw_input("Pick a door number from 1 - %r > " % doorchoice))
if youranswer == correctanswer:
otherremaining = random.choice(incorrectpossibilities)
else:
otherremaining = youranswer
incorrectpossibilities.remove(otherremaining)
print "The correct answer is NOT one of these: %r" % incorrectpossibilities
if youranswer == correctanswer:
print "Which means it is either your answer: %r, or the only other remaining option, %r." % (correctanswer, otherremaining)
else:
print "Which means it is either your answer: %r, or the only other remaining option, %r." % (youranswer, correctanswer)
finalanswer = int(raw_input("Which do you choose? > "))
if finalanswer == correctanswer:
print "You win!"
else:
print "You lose!"
playagain = int(raw_input("Play again? (1 = yes, 0 = no) > "))
print "Thanks for playing!"
gamechoice = int(raw_input("Play one game at a time (1) or play your chosen number of games with one strategy (2)? > "))
playchosengame(gamechoice)
#1 楼
欢迎使用一般的代码审查和编码方法：根据PEP8编写的代码–在您的情况下，这意味着对变量名和函数名使用
snake_case，请在逗号后和运算符之间添加空格。一般而言，您的代码确实看起来不错，这里和那里都有一些小问题，但请通读指南并尝试遵守它们。功能：
incorrect_possibilities = possible_answers使possible_answers无效-稍后再执行incorrect_possiblities.remove(correct_answer)时，这也会更改possible_answers。要获得副本，您需要使用更好的副本版本，或者只是incorrect_possibilities = possible_answers[:]。后者使用切片来表示整个数组，并复制切片区域。使用与
chosen_strategy相关的三元数-根据选择的策略，您可以选择两个选项之一，可以将其编码为final_answer = correct_answer if chosen_strategy == 1 else your_answer，并且在其他情况下也类似。也使用与打印输出有关的三元数–这也可以在打印输出中使用：
print("Which means it is either your answer: {}, or the only other remaining option, {}.".format(
correct_answer if your_answer == correct_answer else your_answer,
other_remaining if your_answer == correct_answer else correct_answer))
使用较新的
print('You chose {}'.format(final_answer)语法–这是Python 2中新代码和Python 3中代码的首选版本。有关示例和说明，请参阅格式规范迷你语言。使用文档字符串
""" ... """，而不是注释描述模块，函数，类和方法时，请参见# ...。 -如果您使用文档字符串，则可以由您的IDE或文档工具使用它们，这取决于Pythonic方式。 :-) 避免使用一次性临时变量–如果使用
playagain来保持游戏的正常进行，并将其结尾更改为while True:，则可以删除if int(raw_input("Play again (0 = no)?")) == 0: break。 （最好在break之前加一个换行符，它确实会打破while循环）可能在主循环上添加循环或验证-如果您在一次玩一个游戏之间进行选择时输入了一些内容，或玩选定数量的游戏，您可以输入错误内容并终止脚本。可以通过在其周围添加一个while循环来增强此功能。
还请注意，如果您输入非数字文本，则代码很容易折断...考虑制作一个通用的int输入函数，该函数在
try ... except函数周围使用input。介绍
if __name__ == '__main__':习惯用法-以前的代码通常用于使代码可作为模块重用，并且在顶层尽可能少地使用代码。结合main()函数，您的代码可轻松重用和扩展。您可以使用（可能通过while循环扩展）：
def main():
game_choice = int(raw_input("Play one game at a time (1) or play your chosen number of games with one strategy (2)? > "))
play_chosen_game(game_choice)
if __name__ == '__main__':
break
请考虑缩短文本长度–您的文本相当有价值，可以从略微缩短文本中受益。并且命名是很好的（单词之间不要使用下划线）。通过学习使用三元运算符（
a if b else c），以及使用与顶级代码和文本长度相关的功能的一点道理，您可以从中受益。#2 楼
阅读BibleStyle Guide查看PEP-8（Python的样式指南）。从您的代码中迅速跳出来的是您有多少个变量和函数
lotsofwordsstucktogetherwithoutanyspacing。这些很难读。首选的Python语法是snake_case。即：def play_chosen_game(game_choice):
...
更容易解析。
Python不像C / Java
不必预先声明变量，也不需要手动递增。您在
one_strategy中的主循环是：i = 1
for i in range(1, num_games + 1):
...
i += 1
这是错误的，原因如下：（1）您不需要初始的
i = 1（2）结尾的i += 1没有意义（3）range()是错误的构造（4）当您确实需要i时，在循环末尾引用num_games。最好选择以下内容：for _ in xrange(num_games):
...
_的命名清楚表明变量并不重要-您只是在做num_games次。另外，我们遍历1..9或0..8都没关系，因此请选择一个更容易拼写的字母。 在此基础上，我们只想计算获胜次数，所以让我们明确地说一下：
wins = 0
for _ in xrange(num_games):
wins += play_game(num_doors, strategy_choice)
移动所有逻辑您已经进入了自己的功能循环。
捡起门
捡起门的逻辑有点难以理解。让我们简化一下。我们从所有门开始：
doors = range(num_doors)
我们随机选择一个正确的门，然后随机选择一个以供选择：
answer = random.randint(0, num_doors - 1)
first_choice = random.randint(0, num_doors - 1)
接下来，根据策略，我们要么留下，要么选择另一扇门：
if strategy_choice == 1:
# pick a new door that isn't first_choice or some random wrong one
goat = random.choice(
[i for i in xrange(num_doors) if i not in (first_choice, answer)]
)
final_answer = random.choice(
[i for i in xrange(num_doors) if i not in (first_choice, goat)]
)
else:
# stay put
final_choice = first_choice
更直接。
#3 楼
对于您编写代码的第二周来说，这是一件好事！我建议阅读Python的官方样式指南PEP8。
我还建议阅读一些我们评价最高的标记了python的问题，以启发人们如何改进/制作东西。
逗号后空格：
在某些情况下，逗号后空格不见了：
for i in range(1,chosennumberofgames + 1):
PEP8建议在逗号后提供一个空格。
snake_case：您应该在
snake_case中命名变量，这也是PEP8所建议的。incorrectpossibilities = possibleanswers
重复逻辑：
我看到很多重复逻辑，例如：
if youranswer == correctanswer:
print "Which means it is either your answer: %r, or the only other remaining option, %r." % (correctanswer, otherremaining)
else:
print "Which means it is either your answer: %r, or the only other remaining option, %r." % (youranswer, correctanswer)
>
我建议使用不同的方法简化这些操作：
answer_to_print_one = your_answer
answer_to_print_two = correct_answer
if your_answer == correct_answer:
answer_to_print_one = correct_answer
answer_to_print_two = other_remaining
print "Which means it is either your answer: {a1}, or the only other remaining option, {a2}.".format(a1=answer_to_print_one, a2=answer_to_print_two)
使用
"" %：这是一个过时且低效的字符串连接方式：
使用
"{string}".format(string=your_string) styl e级联。使用
class：目前，您有一堆随机的关系函数，用于传递输入，参数等。
使用
class代替，这样可以：正确链接功能
在类级别存储参数，减少对参数的需求
存储游戏的多个实例
#4 楼
您的名字很清楚而且很好，但是您的一些名字存在冗长的问题，尤其是冗长的词。例如，def playchosengame(chosengame):
您意识到这是这样的可读性：
def play(chosengame):
当然，您只需用
play(gamechoice)
play_game，但就目前的情况而言，由于参数的命名方式几乎没有歧义。更重要的是，这些行要好读，而不是试图使每个名称告诉您有关其值的所有信息。
评论
您知道没有任何分隔符就很难读很多单词。也许那是值得思考的。下划线_are_the_standard_IIRC。