Python岩石，紙，剪刀遊戲

Question

我正在使用Python，我正在嘗試編寫一個模擬岩石，紙張，剪刀遊戲的簡單程序。除了當我輸入一個無效的回覆（除岩石，紙張或剪刀以外的東西）時，一切都可以正常工作。

Traceback (most recent call last): 
File "C:/Users/home/Desktop/BAGARDNER/Python/rock_pape_scissors.py", line 88, in <module> 
main() 
File "C:/Users/home/Desktop/BAGARDNER/Python/rock_pape_scissors.py", line 14, in main 
number = user_guess() 
File "C:/Users/home/Desktop/BAGARDNER/Python/rock_pape_scissors.py", line 48, in user_guess 
return number 
UnboundLocalError: local variable 'number' referenced before assignment 

我明白，這是告訴我，號碼不被引用，但是從我瞭解的代碼，它不應該需要一個號碼時限定符是假的。

#import random module 
import random 
#main function 
def main(): 
    #intro message 
    print("Let's play 'Rock, Paper, Scissors'!") 
    #call the user's guess function 
    number = user_guess() 
    #call the computer's number function 
    num = computer_number() 
    #call the results function 
    results(num, number) 

#computer_number function 
def computer_number(): 
    #get a random number in the range of 1 through 3 
    num = random.randrange(1,4) 
    #if/elif statement 
    if num == 1: 
     print("Computer chooses rock") 
    elif num == 2: 
     print("Computer chooses paper") 
    elif num == 3: 
     print("Computer chooses scissors") 
    #return the number 
    return num 

#user_guess function 
def user_guess(): 
    #get the user's guess 
    guess = input("Choose 'rock', 'paper', or 'scissors' by typing that word. ") 
    #while guess == 'paper' or guess == 'rock' or guess == 'scissors': 
    if is_valid_guess(guess): 
     #if/elif statement 
     #assign 1 to rock 
     if guess == 'rock': 
      number = 1 
     #assign 2 to paper 
     elif guess == 'paper': 
      number = 2 
     #assign 3 to scissors 
     elif guess == 'scissors': 
      number = 3 
     return number 
    else: 
     print('That response is invalid.') 
     user_guess() 

def is_valid_guess(guess): 
    if guess == 'rock' or 'paper' or 'scissors': 
     status = True 
    else: 
     status = False 
    return status 

def restart(): 
    answer = input("Would you like to play again? Enter 'y' for yes or \ 
    'n' for no: ") 
    #if/elif statement 
    if answer == 'y': 
     main() 
    elif answer == 'n': 
     print("Goodbye!") 
    else: 
     print("Please enter only 'y' or 'n'!") 
     #call restart 
     restart() 

#results function 
def results(num, number): 
    #find the difference in the two numbers 
    difference = num - number 
    #if/elif statement 
    if difference == 0: 
     print("TIE!") 
     #call restart 
     restart() 
    elif difference % 3 == 1: 
     print("I'm sorry! You lost :(") 
     #call restart 
     restart() 
    elif difference % 3 == 2: 
     print("Congratulations! You won :)") 
     #call restart 
     restart() 

main() 

謝謝你的幫忙！

Answer 1

這是你的問題：

if guess == 'rock' or 'paper' or 'scissors': 

這條線is_valid_guess沒有做什麼，你認爲它。相反，它總是返回True。你正在尋找的是這樣的：

if guess == 'rock' or guess == 'paper' or guess == 'scissors': 

或更簡潔：

if guess in ('rock', 'paper', 'scissors'): 

---

的問題是，你有什麼總是返回True因爲Python如何在一個字符串評估布爾上下文。該生產線if guess == 'rock' or 'paper' or 'scissors':計算結果爲：

if (guess == 'rock') or ('paper') or ('scissors'): 

這意味着就是，Python檢查，看看是否guess == 'rock'。如果這是真的，則條件評估爲True。如果它是錯誤的，解釋器會嘗試評估bool('paper')。由於all non-empty strings are "truthy"因此總是評估爲True。因此，您的整個條件總是True，並且每個字符串都是「有效的」。

因此，您的代碼會將所有字符串視爲「有效」，然後在未將實際支持的猜測分配給某個數字時爆炸。

---

最後一點，你is_valid_guess方法可以精簡一點，因爲你只是回到你的布爾表達式的結果。而不是將status變量用作中間變量，只需計算表達式並立即返回即可。我還使用字符串對象的lower()方法來允許不區分大小寫的猜測，以防萬一您要允許。

def is_valid_guess(guess): 
    return guess.lower() in ('rock', 'paper', 'scissors') 

---

你有另外一個問題，你在評論中提到：你在遞歸的方式實現user_guess，所以它如果用戶輸入一個無效的猜測調用自身。但是，在這種情況下，它不會返回遞歸調用的結果。您需要通過更改的user_guess最後一行返回遞歸結果：

return user_guess() 

否則你應該做的是功能使用一個循環，而不是遞歸，這是我會做什麼，因爲功能不固有遞歸。你可以做這樣的事情：

def user_guess(): 
    # get first guess 
    guess = input("Choose 'rock', 'paper', or 'scissors' by typing that word. ") 

    # If that guess is invalid, loop until we get a valid guess. 
    while not is_valid_guess(guess): 
     print('That response is invalid.') 
     guess = input("Choose 'rock', 'paper', or 'scissors' by typing that word. ") 

    # Now assign the (valid!) guess a number 
    # This dictionary is just shorthand for your if/elif chain. 
    guess_table = { 
     'rock' : 1, 
     'paper' : 2, 
     'scissors' : 3 
    } 

    # Return the number associated with the guess. 
    return guess_table[guess.lower()] 

Answer 2

變化

if guess == 'rock' or 'paper' or 'scissors': 

到

if guess == 'rock' or guess == 'paper' or guess == 'scissors': 

事實上，爲了使功能精簡越好，只是這樣做：

def is_valid_guess(guess): 
    return guess == 'rock' or guess == 'paper' or guess == 'scissors' 

Answer 3

其他用戶指出，你需要改變你的驗證在is_valid_guess到：

if guess == 'rock' or guess == 'paper' or guess == 'scissors': 

雖然這不會解決你眼前的問題，這是件好事（給予好評，值得）的意見，並會讓你避免你會遇到一些錯誤。 。

另外，無論什麼用戶輸入，你總是返回他們鍵入爲了防止這種情況，你必須在你的else塊返回user_guess()：

if is_valid_guess(guess): 
    #if/elif statement 
    #assign 1 to rock 
    if guess == 'rock': 
     number = 1 
    #assign 2 to paper 
    elif guess == 'paper': 
     number = 2 
    #assign 3 to scissors 
    elif guess == 'scissors': 
     number = 3 
    return number 
else: 
    print('That response is invalid.') 
    return user_guess() # <-- right here 

Answer 4

只要改變input到raw_input