Python 检查列表中哪些单词可以由一串字母组成

系统：Mac Python 2.7

嗨， 因此，我有一个列表，其中包含我在网上找到的英语词典中的单词。接下来，我有一个由小写字母组成的字符串。我想做的是找到列表（英语词典）中由字符串中的字母组成的所有单词，并将它们保存到单独的列表中。此外，我不希望能够重复使用字母（胶水->欢乐），但给出的字母可以重新排列。示例字符串可以是“hideg”，结果应生成一个包含以下内容的列表：

['hide','hid','hie','dig','eh','he','hi','die','ed','id','i']

以下是我到目前为止得到的（注意：myWord是列表（英语词典），myLetters是字符串）：

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代码中的问题是，您正在比较精确匹配，在大多数情况下，这些字母没有被完全使用，即隐藏！=希德，但它当然可以构成这个词

一种简单的方法（尽管未优化）是利用，如下所示：

In [32]: from collections import Counter

In [33]: words = ['hide','hid','hie','dig','eh','he','hi','die','ed','id','i']

In [34]: letters = 'hide'

In [35]: def valid_word(words, letters):
   ....:     results = []
             # looping through words list and do a comparison
   ....:     for word in words:
                 # build a letter counter for letters
   ....:         letter_counter = Counter(letters)
                 # subtract their letter count, -ve means cannot form the word
   ....:         letter_counter.subtract(Counter(word))
   ....:         if any(c < 0 for c in letter_counter.values()):
   ....:             continue
                 # only append the result if >=0 letters left in counter
   ....:         results.append(word)
   ....:     return results
   ....: 

---

.subtract（）做什么？还有，Counter（）构成字符串的数据类型是什么？@SirPanda您可以阅读我用计数器发布的链接。它基本上是一个统计所有字母计数的工具，

subtract

将用word的字母计数器减去字母计数器，因此如果有-ve计数（这意味着没有足够的字母组成单词）应该被排除在外。

In [32]: from collections import Counter

In [33]: words = ['hide','hid','hie','dig','eh','he','hi','die','ed','id','i']

In [34]: letters = 'hide'

In [35]: def valid_word(words, letters):
   ....:     results = []
             # looping through words list and do a comparison
   ....:     for word in words:
                 # build a letter counter for letters
   ....:         letter_counter = Counter(letters)
                 # subtract their letter count, -ve means cannot form the word
   ....:         letter_counter.subtract(Counter(word))
   ....:         if any(c < 0 for c in letter_counter.values()):
   ....:             continue
                 # only append the result if >=0 letters left in counter
   ....:         results.append(word)
   ....:     return results
   ....: 

In [36]: valid_word(words, letters)
Out[36]: ['hide', 'hid', 'hie', 'eh', 'he', 'hi', 'die', 'ed', 'id', 'i']

In [37]: valid_word(words, 'hideg')
Out[37]: ['hide', 'hid', 'hie', 'dig', 'eh', 'he', 'hi', 'die', 'ed', 'id', 'i']