Question

我正在使用python，我想制作一个刽子手游戏，我做了大部分。玩家1写入一个单词，然后玩家2将通过一个又一个字母来猜测它是什么单词，你总共有7次尝试。我现在唯一需要的东西可以从玩家2猜测的单词中随机选择一个字母。当玩家达到3次以下或类似的尝试时。我真的很新编码，我不知道多少例如我尝试过：if tries＆lt; 3：print（random.letter'wordmaking'）其中wordmaking是输入的名称和玩家1输入的单词。让我们说你和朋友一起玩刽子手而你无法找出他用过的单词或字母。将会有一个代码，为您提供一个尚未猜到的字母：猜猜词：_ _ _ _你剩下4次尝试让我们说这个单词是D E S K并且他输入了字母G，剩下的尝试将下降到3并且游戏将输入Guess字：D _ _ _你还有3次尝试。当他达到一定程度的尝试作为暗示时，它增加了一封信。就像画画一样，只有当你没有多少时间时，他们会给你提示。

Answer 1

>>> from random import choice
>>> some_word = 'wordmaking'
>>> choice(some_word)
'i'
>>> choice(some_word)
'k'
>>> choice(some_word)
'i'
>>> choice(some_word)
'g'

Answer 2

那会对你想要做的事情有用吗？

import random

tries=0 #iterator initiation
if tries<=3:
    letter=random.choice("thewordyouwannaselectlettersfrom")
    print(letter)

else:
    tries+=1
    continue

Answer 3

忍不住写一个刽子手：

import random

# take input for the word of player 1
target = input('Player 1 word: ')

# set guess to empty, but same length as 'target'
guess = '_'*len(target)
tries = 0

while (guess != target) and (tries < 3):
    # take input of player 2
    s = input('Now: "' + guess + '"; Player 2 can guess: ')

    # if guessed input is longer than 1 character, skip it
    if len(s) > 1:
        print('only 1 character allowed at a time')
        continue 

    # update guess with guessed letter if it hits
    guess = ''.join(s if target[i] == s else guess[i] for i in range(len(target)))

    # test if there were any correct guesses, otherwise tries += 1
    if not any(target[i] == s for i in range(len(target))):
        tries+=1

# print result after exiting loop
print('Player {} WINS: word was {}'.format('2' if tries < 2 else '1', target))

蟒蛇。来自特定单词的随机字母

3 个答案: