如何避免重复字符串？

Question

我的代码是：

import random
WORDS = ('python', 'football', 'facebook', 'photo') #list of words that will be riddled
word = random.choice(WORDS)
correct = word
jumble = ''
hint = 'hint'
score = 0
while word:
    position = random.randrange(len(word))
    jumble += word[position] 
    word = word[:position] + word[(position + 1):] #creating jumble of correct words
print('Welcome to the game "Anagrams"')
print('Here`s your anagram:', jumble) #Welcoming and giving a jumble to a player
guess = input('\nTry to guess the original word: ')
if guess == correct:
    score += 5
    print('You won! Congratulations!') #end of game in case of right answer
if guess == hint: #situation if player asks a hint
    if correct == WORDS[0]:
        print('snake')
    elif correct == WORDS[1]:
        print('sport game')
    elif correct == WORDS[2]:
        print('social network')
    elif correct == WORDS[3]:
        print('picture of something')
    score += 1
while guess != correct and guess != '': #situation if player is not correct
    print('Sorry, you`re wrong :(')
    guess = input('Try to guess the original word: ')
print('Thank you for participating in game.')
print('Your score is', score)
input('\nPress Enter to end')

当询问提示字符串时：

“对不起，你错了:(”

重复。
看起来像：

尝试猜测原始单词：提示
体育游戏
对不起，你错了:(

如何使此字符串仅在错误猜测的情况下出现？

Answer 1

最后一次将您更改为：

while guess != correct and guess != '':
    guess = input("Sorry, you`re wrong:( ")

Answer 2

在您的代码中，当播放器类型hint时，播放器会得到提示，但是程序会根据correct单词测试'hint'字符串。 当然， 'hint'不是正确的答案，因此您的程序会告诉他们这是错误的。

只是为了好玩，我对您的代码进行了一些优化，并改善了评分逻辑。 :)

您的字母混杂for循环非常聪明，但是使用random.shuffle函数有一种更有效的方法。 此功能将列表随机排序。 因此，我们需要将所选单词转换为列表，将其洗牌，然后将列表重新组合为字符串。

我也替换了您的提示逻辑。 无需进行大量的if测试来查看哪个提示与当前单词一起使用，将每个单词及其相关的提示存储为元组要简单得多。

import random

#Words that will be riddled, and their hints
all_words = (
    ('python', 'snake'),
    ('football', 'sport game'),
    ('facebook', 'social network'),
    ('photo', 'picture of something'),
)

#Randomly choose a word
word, hint = random.choice(all_words)

#Jumble up the letters of word
jumble = list(word)
random.shuffle(jumble)
jumble = ''.join(jumble)    

print('Welcome to the game "Anagrams"\n')
print('You may ask for a hint by typing hint at the prompt')
print('Wrong guesses cost 2 points, hints cost 1 point\n')

print("Here's your anagram:", jumble)

score = 0
while True:
    guess = input('\nTry to guess the original word: ')
    if guess == word:
        score += 5
        print('You won! Congratulations!')
        break

    if guess == 'hint':
        #Deduct a point for asking for a hint
        score -= 1
        print(hint)
        continue

    #Deduct 2 points for a wrong word
    score -= 2
    print('Sorry, you`re wrong :(')

print('Thank you for participating in game.')
print('Your score is', score)
input('\nPress Enter to end')

Answer 3

让我们尝试解决一些问题：

这个

if guess == hint: #situation if player asks a hint

应该是

elif guess == hint: #situation if player asks a hint

而且这对我来说似乎是错误的

while guess != correct and guess != '': #situation if player is not correct
    print('Sorry, you`re wrong :(')
    guess = input('Try to guess the original word: ')

应该更改为（缩进很重要）：

    guess = input('Try to guess the original word: ')
if guess != correct and guess != '': #situation if player is not correct
    print('Sorry, you`re wrong :(')

我没有在完整的程序中尝试过此更正。

Answer 4

正确猜测和特殊输入"hint"特殊逻辑仅在第一次猜测时运行一次。 之后，您的错误值循环将始终运行。 我认为您想将所有逻辑移入循环：

while True:  # loop forever until a break statement is reached
    guess = input('\nTry to guess the original word: ')
    if guess == correct:
        score += 5
        print('You won! Congratulations!')
        break # stop looping
    if guess == hint: # special case, asking for a hint
        if correct == WORDS[0]:
            print('snake')
        elif correct == WORDS[1]:
            print('sport game')
        elif correct == WORDS[2]:
            print('social network')
        elif correct == WORDS[3]:
            print('picture of something')
        score += 1
    else: #situation if player is not correct, and not askng for a hint
        print('Sorry, you`re wrong :(')

我省去了您的代码将在空输入中退出循环的情况。 如果需要，您应该使用break语句显式添加它作为额外的情况。