Question

我正在制作类似于python上的文字扭曲的游戏，我想知道如何防止用户能够输入两次相同的单词。这是我到目前为止所拥有的......

import random
correct = 0
incorrect = 0
usedwords = []
print 'Welcome to text twist, you have 14 guesses to get 7 words made up of 4, 5 or 6 letters. Good Luck!'

for i in range(14):
    print "Your letters are 'E' 'P' 'L' 'B' 'E' 'B', what is your guess?"
    answer = raw_input()
    if answer in usedwords:
        print "Sorry, you've already used this word"
    if answer == 'belle' or answer == 'bleep' or answer == 'pebble' or answer == 'beep' or answer == 'bell' or answer == 'peel' or answer == 'peep':
        if answer in usedwords:
            print 'Nice that was one of the words!'
        usedwords.append(answer)
        correct = correct + 1
    if answer != 'belle' and answer != 'bleep' and answer != 'pebble' and answer != 'beep' and answer != 'bell' and answer != 'peel' and answer != 'peep':
        print 'Sorry, that was not one of the words.'
        incorrect = incorrect + 1


print 'Your final score was', correct, 'correct and', incorrect, 'wrong.'

Answer 1

import random
correct = 0
incorrect = 0
usedwords = []
print 'Welcome to text twist, you have 14 guesses to get 7 words made up of 4, 5 or 6 letters. Good Luck!'

for i in range(14):
    print "Your letters are 'E' 'P' 'L' 'B' 'E' 'B', what is your guess?"
    answer = raw_input()
    if answer in usedwords:
        print "Sorry, you've already used this word"
    else: #Newly added
        usedwords.append(answer) #We have to add the input words to the list if they are new
        if answer == 'belle' or answer == 'bleep' or answer == 'pebble' or answer == 'beep' or answer == 'bell' or answer == 'peel' or answer == 'peep':
            if answer in usedwords:
                print 'Nice that was one of the words!'
            usedwords.append(answer)
            correct = correct + 1
        if answer != 'belle' and answer != 'bleep' and answer != 'pebble' and answer != 'beep' and answer != 'bell' and answer != 'peel' and answer != 'peep':
            print 'Sorry, that was not one of the words.'
            incorrect = incorrect + 1


print 'Your final score was', correct, 'correct and', incorrect, 'wrong.'

Answer 2

此代码可以修改如下：

import random
correct = 0
incorrect = 0
usedwords = []
print 'Welcome to text twist, you have 14 guesses to get 7 words made up of 4, 5 or 6 letters. Good Luck!'
for i in range(14):
    print "Your letters are 'E' 'P' 'L' 'B' 'E' 'B', what is your guess?"
    answer = raw_input()
    if answer in usedwords:
        print "Sorry, you've already used this word"
        incorrect = incorrect + 1
        continue
    elif answer == 'belle' or answer == 'bleep' or answer == 'pebble' or answer == 'beep' or answer == 'bell' or answer == 'peel' or answer == 'peep':
        print 'Nice that was one of the words!'
        usedwords.append(answer)
        correct = correct + 1
    else:
        print 'Sorry, that was not one of the words.'
        incorrect = incorrect + 1


print 'Your final score was', correct, 'correct and', incorrect, 'wrong.'

<强>的变化：

包含“Continue”语句，以便在同一个单词时跳过迭代重复。
不正确=错误+ 1，在重复同一个字时增加不正确的尝试。
使用if..elif..else而不是多个if语句以更清晰，并忽略“！=”验证。
删除了“使用过的词中的iList itemf answer：”在第二个if语句中进行验证，因为之前已经验证过。
如果单词不在用词中，或者单词不在预期单词列表中，那么这个单词显然不正确，因此在else语句中，错误计数会随着消息而增加。

<强>建议：您可以声明一个列表，而不是在elif语句中使用多个OR， expectedwords = ['belle'，'bleep'，'pebble'，'beep'，'bell'，'peel'，'peep'] 并使用，预期词中的elif答案： ....

阻止用户两次输入相同的输入

2 个答案: