解析电子邮件以识别关键字

这是一个解决你问题的简单方法。我发现使用lambda语法比嵌套列表理解更容易阅读。

keywords = ['information', 'boxes']

sentences = [['this is a paragraph there should be lots more words here'],
             ['more information in this one'],
             ['just more words to be honest, not sure what to write']]


results_lambda = list(
    filter(lambda sentence: any((word in sentence[0] for word in keywords)), sentences))

print(results_lambda)

[['more information in this one']]

这可以通过快速列表理解来完成!

lists = [['here is one sentence'], ['and here is another'], ['let us filter!'], ['more than one word filter']] filter = ['filter', 'one'] result = list(set([x for s in filter for x in lists if s in x[0]])) print(result)

[['let us filter!'], ['more than one word filter'], ['here is one sentence']] 希望这有帮助!

是否要查找包含关键字列表中所有单词的句子?

如果是这样,那么您可以使用一组关键字,并根据列表中是否存在所有单词来过滤每个句子:

keyword_set = set(keywords)
n = len(keyword_set) # number of keywords
def allKeywdsPresent(sentence):
    return len(set(sentence.split(" ")) & keyword_set) == n # the intersection of both sets should equal the keyword set

filtered = [sentence for sentence in sentences if allKeywdsPresent(sentence)]

# filtered is the final set of sentences which satisfy your condition

# if you want a list of booleans:
boolean_array = [allKeywdsPresent(sentence[0]) for sentence in sentences]

另外,要明白使用集合意味着关键字列表中的重复项将被消除。因此,如果您有一个包含一些重复的关键字的列表,那么使用dict而不是set来记录每个关键字的数量,并重用上述逻辑。

从您的示例来看,至少有一个关键字匹配就足够了。然后需要修改allKeywdsPresent()

def allKeywdsPresent(sentence):
   return any(word in keyword_set for word in sentence.split()) 

如果你只想匹配整个单词,而不只是子字符串,你必须考虑所有的单词分隔符(空格、puctuation等),首先将句子拆分成单词,然后将它们与关键词匹配。最简单的,尽管不是傻瓜式的方法是只使用正则表达式 \W

一旦你有了文本中的单词列表和要匹配的关键字列表,查看是否存在匹配的最简单、可能也是最有效的方法就是在两者之间设置交集。因此:

# not sure why you have the sentences in single-element lists, but if you insist...
sentences = [['this is a paragraph there should be lots more words here'],
             ['more information in this one'],
             ['just more disinformation, to make sure we have no partial matches']]

keywords = {'information', 'boxes', 'porcupine'}  # note we're using a set here!

WORD = re.compile(r"\W+")  # a simple regex to split sentences into words

# finally, iterate over each sentence, split it into words and check for intersection
result = [s for s in sentences if set(WORD.split(s[0].lower())) & keywords]
# [['more information in this one']]

那么,它是如何工作的呢?很简单,我们迭代每一个句子(并用小写字母表示大小写不敏感),然后用前面提到的正则表达式将句子拆分成单词。这意味着,例如,第一句话将分为:

['this', 'is', 'a', 'paragraph', 'there', 'should', 'be', 'lots', 'more', 'words', 'here']

然后我们将其转换为一个集合,以便进行快速比较( set 是一个散列序列,基于散列的求交速度非常快),并且,作为一个额外的功能,这还可以消除重复的单词。

最后,我们做的是与我们的 keywords -如果返回任何内容,则这两个集合至少有一个相同的单词,这意味着 if ... 比较评估为 True 在这种情况下,当前句子会添加到结果中。

\W+ nltk