python - 如何不将整个单词 "king"匹配到 "king?"？

Question

如何验证字符串中是否出现了确切的单词？

我需要考虑以下情况，例如“king”之类的词紧跟问号，如下例所示。

unigrams 这应该是 False

In [1]: answer = "king"
In [2]: context = "we run with the king? on sunday"

n_grams 这应该是 False

In [1]: answer = "king tut"
In [2]: context = "we run with the king tut? on sunday"

unigrams 这应该是 True

In [1]: answer = "king"
In [2]: context = "we run with the king on sunday"

n_grams 这应该是 True

In [1]: answer = "king tut"
In [2]: context = "we run with the king tut on sunday"

正如人们提到的，对于 unigram 的情况，我们可以通过将字符串拆分为列表来处理它，但这对 n_grams 不起作用。

阅读一些帖子后，我认为我应该尝试使用后视来处理，但我不确定。

Answer 1

return answer in context.split():

>>> answer in context.split()
False

你不需要正则表达式。

如果您正在寻找关键字:

all([ans in context.split() for ans in answer.split()])

将与 "king tut" 一起使用，但这取决于您是否要匹配如下字符串:

"we tut with the king"

如果不这样做，您仍然需要正则表达式 (although you should probably use one) ，假设您只想考虑整个术语(默认情况下通过 .split() 正确拆分):

def ngram_in(match, string):
    matches = match.split()
    if len(matches) == 1:
        return matches[0] in string.split()
    words = string.split()
    words_len = len(words)
    matches_len = len(matches)
    for index, word in enumerate(words):
        if index + matches_len > words_len:
            return False
        if word == matches[0]:
            for match_index, match in enumerate(matches):
                potential_match = True
                if words[index + match_index] != match:
                    potential_match = False
                    break
            if potential_match == True:
                return True
    return False

这是 O(n*m) 在最坏情况下的字符串，大约是正则表达式在正常字符串上的一半。

>>> ngram_in("king", "was king tut a nice dude?")
True
>>> ngram_in("king", "was king? tut a nice dude?")
False
>>> ngram_in("king tut a", "was king tut a nice dude?")
True
>>> ngram_in("king tut a", "was king tut? a nice dude?")
False
>>> ngram_in("king tut a", "was king tut an nice dude?")
False
>>> ngram_in("king tut", "was king tut an nice dude?")
True