Question

我目前对python和编程世界并不陌生，所以这个问题对您来说可能很新。

我正在尝试创建一个石头，纸张，剪刀，蜥蜴和Spock游戏。我想使用模来保存很多if语句。我的问题是我无法以取模来确定正确的获胜者。有人可以看到我在做什么错吗？

def syracuse(nr, seq=[]):
nr = int(nr)
if nr != 1:
    if (nr % 2 != 0):
        n =  3*nr+1
    else:
        n = nr // 2
    seq.append(n)
    print(seq)
    syracuse(n, seq)
else:
    return seq

例如，当我输入import random while True: User1 = input('What s your name?>>>') print("Lets start a single player mode of rock, paper, scissors, lizard and Spock", end='.') print('\n choose one of the options below', User1, '(number only):''\n 0.rock \n 1.paper \n 2.scissors \n 3.lizard \n 4.Spock') player_num = int(input()) computer_num = random.randrange(0, 5) difference = (player_num - computer_num) % 5 if player_num == -5: winner = 'You lost' else: if difference == 0: winner = 'It´s a tie' print(computer_num) elif difference == 1 or difference == 2: winner = (User1 + "wins!") print(computer_num) elif difference == 3 or difference == 4: winner = 'Computer wins.' print(computer_num) else: print("Please select one of the options followed by 1, 2, 3 or 4.") print(winner)时，计算机选择2并赢得比赛。

Answer 1

我建议将“从用户那里获取号码”放入函数中，并为愚蠢的玩家添加一些异常处理：

import random

def get_numeric_input(text,num_range):
    i = -1
    while i < 0:
        k = input(text)
        try:
            i = int(k)
            if i in num_range:
                return i
            else:
                i = -1
                raise ValueError
        except ValueError:
            print("Only numbers in range: ",list(num_range))

您可以在Asking the user for input until they give a valid response的答案中了解更多有关此的信息。

您只需要向用户询问一次他的名字，然后将其余游戏逻辑放入带有适当条件的while中即可保留-我选择输掉5次就足够了。

收集赢/抽/输成变量。我为选择映射添加了一个用于数字的字典，并为更好的输出添加了一些string.format（）-请参阅：

User1 = input('What s your name?>>>')

print("Lets start a single player mode of rock, paper, scissors, lizard and Spock.")
lost = 0
win = 0
tie= 0

# dictionary lookup for nicer outputs - maps number to text
d = { 0:"rock", 1:"paper", 2:"scissors", 3:"lizard", 4:"Spock"}

# define some texts once if they repeat all the time
rules = ('''
Choose one of the options below {} - use number only:
    0. rock 
    1. paper
    2. scissors
    3. lizard
    4. Spock
Your choice? ''')
userwins = User1 + " wins!" 

while lost < 5:

    # fully automatic play - use the commented instead for human players
    player_num = random.randrange(0, 5) # get_numeric_input(rules,range(5))
    computer_num = random.randrange(0, 5)

    # always print what was choosen
    print("You took {} - the computer choose {}.".format(d[player_num], d[computer_num]), 
          end=" " )

    difference = (player_num - computer_num) % 5
    if difference == 0:
        tie += 1
        print('It´s a tie') 
    elif difference in (1,2):
        win += 1
        print(userwins)
    else:
        lost += 1
        print('Computer wins.')

print("You won {} and lost {} times. You tied {} times. Overall you are the {}.".format(
    win, lost, tie, "Winner" if win > lost else "Looser" ))

对于实际的游戏逻辑-您的映射似乎不正确。游戏规则如下：

def did_I_win(me,pc):
    # 0 rock     = scissor, lizard
    # 1 paper    = rock, spock
    # 2 scissor  = lizard, paper
    # 3 lizard   = paper, spock
    # 4 spock    = rock, scissor
    wins = {0:{2,3}, 1:{0,4},2:{1,3},3:{1,4},4:{0,2}}
    return pc in wins[me]

因此，如果您将主程序更改为：

while lost < 5:

    # fully automatic play - use the commented instead for human players
    player_num = random.randrange(0, 5) # get_numeric_input(rules,range(5))
    computer_num = random.randrange(0, 5)

    # always print what was choosen
    print("You took {} - the computer choose {}.".format(d[player_num], d[computer_num]), 
          end=" " )

    difference = (player_num - computer_num) % 5
    if player_num == computer_num:
        tie += 1
        print('It´s a tie') 
    elif did_I_win(player_num,computer_num):
        win += 1
        print(userwins)
    else:
        lost += 1
        print('Computer wins.')

print("You won {} and lost {} times. You tied {} times. Overall you are the {}.".format(
    win, lost, tie, "Winner" if win > lost else "Looser" ))

它按预期工作：

What s your name?>>>Joe
Lets start a single player mode of rock, paper, scissors, lizard and Spock.
You took Spock - the computer choose rock. Joe wins!
You took rock - the computer choose Spock. Computer wins.
You took paper - the computer choose scissors. Computer wins.
You took paper - the computer choose paper. It´s a tie
You took paper - the computer choose lizard. Computer wins.
You took rock - the computer choose Spock. Computer wins.
You took rock - the computer choose Spock. Computer wins.
You won 1 and lost 5 times. You tied 1 times. Overall you are the Looser.

您当前的逻辑为您提供：

                                           Result   ok? 
rock       vs paper      = (0-1) % 5 = 4   lost     ok
rock       vs scissors   = (0-2) % 5 = 3   win      ok
rock       vs lizard     = (0-3) % 5 = 2   win      ok
rock       vs Spock      = (0-4) % 5 = 1   lost     ok
paper      vs rock       = (1-0) % 5 = 1   win      ok
paper      vs scissors   = (1-2) % 5 = 4   lost     ok
paper      vs lizard     = (1-3) % 5 = 3   lost     no: earlier 3 was win
paper      vs Spock      = (1-4) % 5 = 2   win      ok
scissors   vs rock       = (2-0) % 5 = 2   lost     no: earlier 2 was win
scissors   vs paper      = (2-1) % 5 = 1   ... etc ...
scissors   vs lizard     = (2-3) % 5 = 4
scissors   vs Spock      = (2-4) % 5 = 3
lizard     vs rock       = (3-0) % 5 = 3
lizard     vs paper      = (3-1) % 5 = 2
lizard     vs scissors   = (3-2) % 5 = 1
lizard     vs Spock      = (3-4) % 5 = 4
Spock      vs rock       = (4-0) % 5 = 4
Spock      vs paper      = (4-1) % 5 = 3
Spock      vs scissors   = (4-2) % 5 = 2

程序：

d= { 0:"rock",1:"paper",2:"scissors",3:"lizard",4:"Spock"}

# rock = scissor, lizard
# paper = rock, spock
# scissor = lizard, paper
# lizard= paper, spock
# spock = rock, scissor
print("------")
for i in range(5):
    for j in range(5):
        if i!=j: 
            print("{:<10} vs {:<10} = ({}-{}) % 5 = {}".format(
                d[i],d[j],i,j,(i-j)%5))

岩石，纸剪刀Spock蜥蜴python，具有模算法

1 个答案: