Python 3.x 石头、纸、剪刀蟒蛇
我正在学习python并编写一个石头、纸和剪刀游戏。我可以运行该程序,但每次程序通过嵌套的if、elif、else语句时,无论humplayer变量的值是什么,输出都将继续是外部的else语句。有什么帮助吗 谢谢大家!Python 3.x 石头、纸、剪刀蟒蛇,python-3.x,Python 3.x,我正在学习python并编写一个石头、纸和剪刀游戏。我可以运行该程序,但每次程序通过嵌套的if、elif、else语句时,无论humplayer变量的值是什么,输出都将继续是外部的else语句。有什么帮助吗 谢谢大家! import random def main(): playerName=introduction() humplayer=hum_player() compchoice=com_player() tieScore = 0 hum
import random
def main():
playerName=introduction()
humplayer=hum_player()
compchoice=com_player()
tieScore = 0
humScore=0
comScore=0
humplayer=input("Enter your move (0 for Rock, 1 for Paper, and 2 for Scissors): ")
while humplayer != -1:
compchoice=com_player()
result= evaluate_Game(humplayer,compchoice,playerName)
if result==0:
print("It's a tie!")
tieScore+=1
elif result==1:
comScore+=1
else:
humScore+=1
humplayer=input("Enter your move (0 for Rock, 1 for Paper, and 2 for Scissors): ")
print(statistics(playerName,humScore,comScore,tieScore))
def introduction():
print("Welcome to the game of Rock, Paper, Scissors. You will be playing against the computer.")
name= input("What is your name? ")
print("Here are the rules", name+":")
print(" If a player chooses Rock and the other chooses Scissors, Rock wins.")
print(" If a player chooses Scissors and the other chooses Paper, Scissors wins.")
print(" If a player chooses Paper and the other chooses Rock, Paper wins.")
print(" If both players make the same choice, it's a tie.")
print(" Enter -1 to quit the game ")
return name
def hum_player():
choice = int(input("Enter your move (0 for Rock, 1 for Paper, and 2 for Scissors): "))
return choice
def com_player():
random_Num = random.randint(0,2)
return(random_Num)
def evaluate_Game(humplayer,compchoice,playerName):
a = "Rock"
b="Paper"
c="Scissors"
if humplayer==0:
if compchoice==0:
return 0
elif compchoice==1:
print(playerName, "plays ",a," computer plays",b)
print("Paper covers Rock, Computer wins!")
return 1
else:
print(playerName, "plays",a,"computer plays", c)
print("Rock crushes Scissors ,", playerName," wins!")
return 2
elif humplayer==1:
if compchoice==0:
print(name,"plays",b," computer plays", a)
print("Paper covers Rock.", playerName,"wins!")
return 2
elif compchoice==1:
print(playerName,"plays", b," computer plays" , b)
print("It's a tie!")
return 0
else:
print(playerName, "plays", b,"computer plays", c)
print("Scissors cuts Paper. Computer wins!")
return 1
else:
if compchoice==0:
print(playerName, "plays", c," computer plays", a)
print("Rock breaks Scissors. Computer wins!")
return 1
elif compchoice==1:
print(playerName, "plays", c, " computer plays", b)
print("Scissors cuts Paper." , playerName, "wins!")
return 2
else:
print(playerName, "plays", c," computer plays", c)
print("It's a tie!")
return 0
def statistics(playerName,humScore,tieScore,comScore):
print("There were", tieScore+comScore+humscore, "games:", playerName, "won", humScore, "games, the computer won", comScore, "games and there were", tieScore, "ties.")
main()
当你向人类请求输入时,你并没有将它转换为
int
。而不是重写这一行:
humplayer=input("Enter your move (0 for Rock, 1 for Paper, and 2 for Scissors): ")
为什么不呢
如果你不使用你编写的函数
humplayer = hum_player()
虽然在hum\u player()
函数中正确地强制转换为int
,但在所有其他重新查询的地方都不会。你不妨重复使用这个功能。这就是我们首先编写单独函数的原因(因此我们不必重复已经编写的代码)
第二件有点错误的事情是您使用statistics()
方法的方式
def statistics(playerName,humScore,tieScore,comScore):
print("There were", tieScore+comScore+humscore, "games:", playerName, "won", humScore, "games, the computer won", comScore, "games and there were", tieScore, "ties.")
如前所述,这实际上还可以。问题是你用的是
'print(statistics(...))`
实际的statistics
方法返回None
,因此这个print语句将打印None
(尽管实际的函数调用仍将打印到控制台)。更好的方法是调用以下方法之一:
statistics(...)
不使用print语句,或者将您的statistics方法更改为返回字符串而不是打印它
def statistics(playerName,humScore,tieScore,comScore):
return ("There were", tieScore+comScore+humscore, "games:", playerName, "won", humScore, "games, the computer won", comScore, "games and there were", tieScore, "ties.")
感谢在发布这个问题后,我意识到我的输入中没有包含int(),但是统计方法的问题对我来说是个新问题!