python - 在文字游戏中查找单词

Question

我正在尝试编写一个程序，允许用户输入一个单词，然后在单词文本文件中找到隐藏在该单词中的所有长度为 4 或更大的单词。到目前为止，我的代码可以检测到用户输入的单词中没有混淆的单词。例如，如果我输入“houses”，输出将显示“house, houses, ho, us, use, uses”。它还应该识别“软管、软管、鞋、鞋、色调等”。

我知道 itertools 是最简单的解决方案，但我想使用一种仅使用循环、字典和列表的不同方法。

到目前为止，这是我的代码:

def main():
    filename = open('dictionary.txt').readlines()
    word_list = []
    for line in filename:
        word_list.append(line.strip())

    print 'Lets Play Words within a Word!\n'
    word = raw_input('Enter a word: ')
    words_left = 0
    for words in word_list:
        letters = list(words)
        if words in word:
            print words
            words_left += 1
        else:
            False

我尝试创建的输出格式应该如下所示:

Lets play Words within a Word!

Enter a word: exams

exams ---  6 words are remaining
> same #user types in guess
Found!  # prints 'Found!' if above word is found in the dictionary.txt file

exams ---  5 words are remaining
> exam
Found!

exams ---  4 words are remaining
> mesa
Found!

exams ---  3 words are remaining
> quit() #if they type this command in the game will end

我还想添加一个游戏摘要，根据 Scrabble 的字母评分来跟踪用户的分数，但那是稍后的事情。

Answer 1

真正能帮到你的是itertools.permutations .这个方便的函数接受一个序列(例如字符串)并为您提供指定长度的所有排列。

以下是我建议您使用它的方式:

import itertools

def main():
    with open('dictionary.txt') as file:
        word_list = set(line.strip() for line in file) # Use a set instead of a list for faster lookups

    print 'Lets Play Words within a Word!\n'
    word = raw_input('Enter a word: ')

    subwords = set() # In case a word can be made in multiple ways
    for i in range(4, len(word)+1):
        for permutation in itertools.permutations(word, i):
            word_to_check = ''.join(permutation)
            if word_to_check in word_list:
                subwords.add(word_to_check)

这会检查单词的所有可能排列以查看哪些是单词，并保留在单词中找到的所有单词的集合(无重复项)。然后，当用户猜测时，您可以检查

if user_guess in subwords and user_guess not in already_found_words: