我在Python里的第一个Hangman游戏

Question

这是我的第一场比赛，我真的很自豪，但我认为这是可怕的混乱。有什么办法可以改进吗？

#HANGMAN

# Valid Word Checker

def valid_word(word):

    ill_chars = ["!","@","#","$","%","^","&","*","(",")",",",".", " ","1","2","3","4","5","6","7","8","9","0"]
    for i in word:
        if i in ill_chars:
            print(f"\nError: It must be a letter. No symbols, numbers or spaces allowed. \n\n '{i}' is not allowed\n")
            return False
    if len(word) > 12:
        print("\n\nError: Word is too long. Use a word with eight characters or less.")
        return False
    return True

word_list = []
spaces = []
guessed_letters = []

ill_chars = ["!","@","#","$","%","^","&","*","(",")",",",".", " ","1","2","3","4","5","6","7","8","9","0"]

head ="    O"
armL = "   /"
torso = "|" 
armR = "\\"
legL = "/"
legR = " \\"

hangman_parts = [head, armL, torso, armR, legL, legR]
hangman_progress = ["", 
                    "|", 
                  "\n|", 
                  "\n|"]

hangman_final = ["|~~~~|\n"]

# Check if word is valid

wordisValid = False
while True:
    word = input("\n\n\nChoose any word\n\n\n").lower()
    if valid_word(word) == True:
        wordisValid = True
        break
    else:
        continue

# Add to list
for i in word:

    word_list.append(i)
    spaces.append("_ ") 

# Main Game Loop
bad_letter = 0

while wordisValid == True:  


    print("".join(hangman_final))
    print("\n\n\n\n")
    print("".join(spaces))
    print("\n\nThe word has: " + str(len(word)) + " letters.")
    print("\nYou've tried the following letters: " + "\n\n" + "".join(guessed_letters) + "\n\n")


    # Winning Loop

    if "".join(spaces) == word:
        print(f"YOU WIN! The word was: {word}")
        break
    # Choose Letters

    player_guess = input("\n\nPlease choose a letter: \n\n\n\n").lower()
    guessed_letters.append(" " + player_guess)

    if player_guess in ill_chars:
        print(f"\nError: It must be a letter. No symbols, numbers or spaces allowed. \n\n '{player_guess}' is not allowed\n")
    elif len(player_guess) > 1:
        print("\nError: You must use one letter.\n")
    elif player_guess == "":
        print("\nError: No input provided.\n")


    # Wrong Letter
    elif player_guess not in word_list:

        bad_letter += 1

        if bad_letter == 1:
            hangman_final.append(hangman_progress[1] + head)

        elif bad_letter == 2:
            hangman_final.append(hangman_progress[2] + "    " + torso)

        elif bad_letter == 3:
            hangman_final.pop(2)
            hangman_final.append(hangman_progress[2] + armL + torso)

        elif bad_letter == 4:
            hangman_final.pop(2)
            hangman_final.append(hangman_progress[2] + armL + torso + armR)

        elif bad_letter == 5:
            hangman_final.append(hangman_progress[3] + "   " + legL)

        elif bad_letter == 6:
            hangman_final.pop(3)
            hangman_final.append(hangman_progress[3] + "   " + legL + legR)
            print("\n\nThe word was: " + word)
            print("\n\n\n\n\n\n\n" + "".join(hangman_final))
            print(" YOU GOT HUNG ")
            break


        print("".join(hangman_final))
        print("\n\n\n\n")
        print("".join(spaces))
        print("\n\nThe word has: " + str(len(word)) + " letters.")
        print("\nYou've tried the following letters: " + "\n\n" + "".join(guessed_letters) + "\n\n")
        print(f"\n\n\n{player_guess} is not in the word. Try again.\n\n")


    # END GAME
    if bad_letter == 6:
        break

    # Add letters guessed to a list
    counter = 0
    for i in word:

        if player_guess == i:
            spaces[counter] = player_guess
        counter += 1

尽管它是有效的，我不确定我是否有正确的想法，当涉及到作出这种类型的项目。

Answer 1

虽然很明显，您对python还不熟悉，但是仍然可以让它第一次运行。做得好!

输入验证

目前，您有一个禁止字符列表。虽然这样可以工作，但默认情况下python允许unicode输入。这意味着人们可以输入成千上万的字母。我建议使用有效输入的白名单。如下所示：

VALID_LETTERS = "abcdefghijklmnopqrstuvwxyz"  # This won't ever change, and typically this sort
# of things  is a module-level variable. 

def valid_word(word):
    for letter in word:
        if letter not in VALID_LETTERS:
            print(f"Letter {letter} is not a valid letter. Please use just a-z.")
            return
    if len(word) > 8:  # This was 12 in your script. Little mistake? And you might want a 
        # module-level constant for this as well.
        print("\n\nError: Word is too long. Use a word with eight characters or less.")
        return False
    return True

main()和

通常，在python中，我们从不会在加载模块时执行代码。详情请在这里阅读，以及我们如何防范它。我过会儿再谈这个。现在，这意味着我们应该尽可能多地使用函数--让我们称它们为main()和put ()。

main()将是主菜单函数。因为我们没有真正的菜单，这基本上会开始一个游戏，也许会问玩家是否想再玩一次。

游戏()将是一个让我们玩一个游戏的函数。这其中的大部分将是一个循环，以获取玩家的输入和计算游戏状态。基本上，脚本中不属于其他函数的所有内容都应该在game()函数中。

我以后会剪掉一些我们不需要的变量。当我们到达我们使用它们的地方时我会解释的。

如下所示：

def game():
    # Variable naming is important. This is an example of a good name.
    guessed_letters = []

    # ill_chars = [...]  We already have this variable before here!

    # Lets assign the parts directly. We can index them as in "3rd part".
    # And for ourselves, we comment the meaning to make sure we still know what it means a year 
    # from now.
    hangman_parts = [
        "    O",  # Head
        "   /",  # Left Arm
        "|" ,  # Torse
        "\\",  # Right Arm
        "/",  # Left Leg
        " \\"  # Right Leg
    ]
    hangman_progress = ["", 
                        "|", 
                      "\n|", 
                      "\n|"]

    hangman_final = ["|~~~~|\n"]

    # Check if word is valid

    # Naming: should be word_is_valid, to be consistent with your other variables.
    word_is_valid = False
    while not word_is_valid:
        word = input("\n\n\nChoose any word\n\n\n").lower()
        word_is_valid = valid_word(word)
    # With this loop condition, it'll keep asking until valid_word returns True. So we can just 
    # assign that value to word_is_valid directly.

    # Since this is the word we're trying to guess, lets just push it upwards in
    # the console a lot, so we won't see it all the time:
    print("\n" * 100)  # Prints 100 newlines.

    # We don't need word_list, just word will do. We don't need spaces either.
    while True:  # Main game loop. We don't need to check for anything, we'll just use break or
        # return to get out.

现在我们要用一个列表理解。它几乎与生成器表达式相同，但它给出了一个真正的列表，因此我们可以问python它有多长：

        # Calculate how many bad letters we have with a list comprehension:
        bad_letter = len([letter for letter in guessed_letters if letter not in word])
        draw_hangman(bad_letter)
        if bad_letter > 5:
            print("\n\n\n\nYOU GOT HUNG")
            return
        print("\n\n\n\n")

你过去常常在两个不同的地方画你的绞刑架，而在一个地方做则更实用。它还确保如果您想要更改某些内容，则只需更改一次。

我也改变了画它的方法。为这个绘图操作设置一个函数更有用，并让它计算到目前为止如何从坏字母的数量中绘制。这使得绘图独立于当前的游戏状态。下面是我根据您对hangman_final的修改提出的版本：

def draw_hangman(bad_letter):
    print(hangman_final, end="")
    if bad_letter > 0:  # Head
        print(hangman_progress[1] + hangman_parts[0], end="")
    if bad_letter == 2:  # Torso and arms
        print(hangman_progress[2] + "    " + hangman_parts[2], end="")
    elif bad_letter == 3:
        print(hangman_progress[2] + "".join(hangman_parts[1:3]), end="")
    elif bad_letter > 3:
        print(hangman_progress[2] + "".join(hangman_parts[1:4]), end="")
    if bad_letter > 4:  # Legs
        print(hangman_progress[3] + "   " + hangman_parts[4], end="")
    if bad_letter > 5:
        print(hangman_parts[5], end="")

在这里，我们需要打印我们猜到的字母，而下划线不是这样。我们可以在一行中计算和打印：

        print(" ".join(letter for letter in word if letter in guessed_letters else "_"))

对于每个字母，如果它包含在guessed_letters中，它就会打印出来，否则就会打印一个下划线。这是一个生成器表达式。这个函数几乎等于以下函数：

def make_word(word, guessed_letters):
    result = []
    for letter in word:
        if letter in guessed_letters:
            result.append(letter)
        else:
            result.append("_")
    return result

我说这几乎是因为它不完全是从其中产生的列表，即使在for循环中使用它或将它提供给"".join()这样的函数时，它的作用也是一样的。它的作用也更像一个循环，而不是一个函数，但这一点现在并不重要。

我们将结果提供给连接。请注意，我们将其输入到“".join()，中间有一个空格，这将在每个字母之间放一个空格，就像您以前做的那样。

        print(f"\n\nThe word has: {len(word)} letters.")  # f-strings are shorter, and easy to use. 
        print(f"\nYou've tried the following letters:\n\n{''.join(guessed_letters) }\n\n")
        # For a string literal inside an f-string, use the other type of quotes

        # This is another generator expression - this one returns a not-quite-list of booleans, and
        # the all() builtin function returns True if and only if all results of the generator
        # expression are True.
        if all(letter in guessed_letters for letter in word):
            print(f"You WIN!\nThe word was: {word}")
            return

        input_valid = False
        while not input_valid:
            new_letter = input("\n\nPlease choose a letter: \n\n\n\n").lower()
            # VERY GOOD that you lowercase it! Prevents tons of weird behaviour.
            if len(new_letter) > 1:
                print("Please enter a single letter only.")
            elif new_letter not in VALID_LETTERS:
                print(f"{new_letter} is not a valid letter. Please enter a letter a-z")
            elif not new_letter:  # Empty strings evaluate to False
                print("Please enter a letter before pressing enter.")
            elif new_letter in guessed_letters:
                print("You already guessed {new_letter} before!")
            else:
                guessed_letters.append(new_letter):
                input_valid = True

这与你的投入非常接近。但是我把它放在一个循环中，所以如果我们有一个无效的输入，我们就不用再遍历整个游戏循环了。我使用了一个非常简单的信号布尔值来保持它的运行，但是也许有更优雅的方法来做到这一点。不过，保持简单是永远不会错的。

现在，要玩这个游戏，只需调用game()函数。但也许我们想玩多场系列赛？让我们让我们的main()函数：

def main():
    game()
    while "y" in input("Play again? [Y/n]").lower():
        game()

为什么这是在一个单独的功能？再一次，为了扩展性。也许，我们想从另一个模块导入我们的游戏。如果是这样的话，我们不希望它在导入时运行，而是在调用函数时运行。所以我们用这个守卫结束这个脚本：

if __name__ == "__main__":
    main()

如果我们直接执行这个文件，这将运行main()函数并让我们玩游戏。但是如果我们导入它，它就不会，所以我们可以在需要的时候调用这个函数。