Python - RegEx for splitting text into sentences (sentence-tokenizing) [duplicate]

问题

I want to make a list of sentences from a string and then print them out. I don't want to use NLTK to do this. So it needs to split on a period at the end of the sentence and not at decimals or abbreviations or title of a name or if the sentence has a .com This is attempt at regex that doesn't work.

import re

text = """\
Mr. Smith bought cheapsite.com for 1.5 million dollars, i.e. he paid a lot for it. Did he mind? Adam Jones Jr. thinks he didn't. In any case, this isn't true... Well, with a probability of .9 it isn't.
"""
sentences = re.split(r' *[\.\?!][\'"\)\]]* *', text)

for stuff in sentences:
        print(stuff)

Example output of what it should look like

Mr. Smith bought cheapsite.com for 1.5 million dollars, i.e. he paid a lot for it. 
Did he mind?
Adam Jones Jr. thinks he didn't.
In any case, this isn't true...
Well, with a probability of .9 it isn't.

回答1:

(?<!\w\.\w.)(?<![A-Z][a-z]\.)(?<=\.|\?)\s

Try this. split your string this.You can also check demo.

http://regex101.com/r/nG1gU7/27

回答2:

Ok so sentence-tokenizers are something I looked at in a little detail, using regexes, nltk, CoreNLP. You end up writing your own and it depends on the application. This stuff is tricky and valuable and people don't just give their tokenizer code away. (Ultimately, tokenization is not a deterministic procedure, it's probabilistic, and also depends very heavily on your corpus or domain, e.g. social-media posts vs Yelp reviews vs ...)

In general you can't rely on one single Great White infallible regex, you have to write a function which uses several regexes (both positive and negative); also a dictionary of abbreviations, and some basic language parsing which knows that e.g. 'I', 'USA', 'FCC', 'TARP' are capitalized in English.

To illustrate how easily this can get seriously complicated, let's try to write you that functional spec for a deterministic tokenizer just to decide whether single or multiple period ('.'/'...') indicates end-of-sentence, or something else:

function isEndOfSentence(leftContext, rightContext)

Return False for decimals inside numbers or currency e.g. 1.23 , $1.23, "That's just my $.02" Consider also section references like 1.2.3, European date formats like 09.07.2014, IP addresses like 192.168.1.1, MAC addresses...
Return False (and don't tokenize into individual letters) for known abbreviations e.g. "U.S. stocks are falling" ; this requires a dictionary of known abbreviations. Anything outside that dictionary you will get wrong, unless you add code to detect unknown abbreviations like A.B.C. and add them to a list.
Ellipses '...' at ends of sentences are terminal, but in the middle of sentences are not. This is not as easy as you might think: you need to look at the left context and the right context, specifically is the RHS capitalized and again consider capitalized words like 'I' and abbreviations. Here's an example proving ambiguity which : She asked me to stay... I left an hour later. (Was that one sentence or two? Impossible to determine)
You may also want to write a few patterns to detect and reject miscellaneous non-sentence-ending uses of punctuation: emoticons :-), ASCII art, spaced ellipses . . . and other stuff esp. Twitter. (Making that adaptive is even harder). How do we tell if @midnight is a Twitter user, the show on Comedy Central, text shorthand, or simply unwanted/junk/typo punctuation? Seriously non-trivial.
After you handle all those negative cases, you could arbitrarily say that any isolated period followed by whitespace is likely to be an end of sentence. (Ultimately, if you really want to buy extra accuracy, you end up writing your own probabilistic sentence-tokenizer which uses weights, and training it on a specific corpus(e.g. legal texts, broadcast media, StackOverflow, Twitter, forums comments etc.)) Then you have to manually review exemplars and training errors. See Manning and Jurafsky book or Coursera course [a]. Ultimately you get as much correctness as you are prepared to pay for.
All of the above is clearly specific to the English-language/ abbreviations, US number/time/date formats. If you want to make it country- and language-independent, that's a bigger proposition, you'll need corpora, native-speaking people to label and QA it all, etc.
All of the above is still only ASCII. Allow the input to be Unicode, and things get harder still (and the training-set necessarily must be either much bigger or much sparser)

In the simple (deterministic) case, function isEndOfSentence(leftContext, rightContext) would return boolean, but in the more general sense, it's probabilistic: it returns a float 0.0-1.0 (confidence level that that particular '.' is a sentence end).

References: [a] Coursera video: "Basic Text Processing 2-5 - Sentence Segmentation - Stanford NLP - Professor Dan Jurafsky & Chris Manning" [UPDATE: an unofficial version used to be on YouTube, was taken down]

回答3:

Try to split the input according to the spaces rather than a dot or ?, if you do like this then the dot or ? won't be printed in the final result.

>>> import re
>>> s = """Mr. Smith bought cheapsite.com for 1.5 million dollars, i.e. he paid a lot for it. Did he mind? Adam Jones Jr. thinks he didn't. In any case, this isn't true... Well, with a probability of .9 it isn't."""
>>> m = re.split(r'(?<=[^A-Z].[.?]) +(?=[A-Z])', s)
>>> for i in m:
...     print i
... 
Mr. Smith bought cheapsite.com for 1.5 million dollars, i.e. he paid a lot for it.
Did he mind?
Adam Jones Jr. thinks he didn't.
In any case, this isn't true...
Well, with a probability of .9 it isn't.

回答4:

sent = re.split('(?<!\w\.\w.)(?<![A-Z][a-z]\.)(?<=\.|\?)(\s|[A-Z].*)',text)
for s in sent:
    print s

Here the regex used is : (?<!\w\.\w.)(?<![A-Z][a-z]\.)(?<=\.|\?)(\s|[A-Z].*)

First block: (?<!\w\.\w.) : this pattern searches in a negative feedback loop (?<!) for all words (\w) followed by fullstop (\.) , followed by other words (\.)

Second block: (?<![A-Z][a-z]\.): this pattern searches in a negative feedback loop for anything starting with uppercase alphabets ([A-Z]), followed by lower case alphabets ([a-z]) till a dot (\.) is found.

Third block: (?<=\.|\?): this pattern searches in a feedback loop of dot (\.) OR question mark (\?)

Fourth block: (\s|[A-Z].*): this pattern searches after the dot OR question mark from the third block. It searches for blank space (\s) OR any sequence of characters starting with a upper case alphabet ([A-Z].*). This block is important to split if the input is as

Hello world.Hi I am here today.

i.e. if there is space or no space after the dot.

回答5:

Naive approach for proper english sentences not starting with non-alphas and not containing quoted parts of speech:

import re
text = """\
Mr. Smith bought cheapsite.com for 1.5 million dollars, i.e. he paid a lot for it. Did he mind? Adam Jones Jr. thinks he didn't. In any case, this isn't true... Well, with a probability of .9 it isn't.
"""
EndPunctuation = re.compile(r'([\.\?\!]\s+)')
NonEndings = re.compile(r'(?:Mrs?|Jr|i\.e)\.\s*$')
parts = EndPunctuation.split(text)
sentence = []
for part in parts:
  if len(part) and len(sentence) and EndPunctuation.match(sentence[-1]) and not NonEndings.search(''.join(sentence)):
    print(''.join(sentence))
    sentence = []
  if len(part):
    sentence.append(part)
if len(sentence):
  print(''.join(sentence))

False positive splitting may be reduced by extending NonEndings a bit. Other cases will require additional code. Handling typos in a sensible way will prove difficult with this approach.

You will never reach perfection with this approach. But depending on the task it might just work "enough"...

回答6:

Try this:

(?<!\b(?:[A-Z][a-z]|\d|[i.e]))\.(?!\b(?:com|\d+)\b)

回答7:

I wrote this taking into consideration smci's comments above. It is a middle-of-the-road approach that doesn't require external libraries and doesn't use regex. It allows you to provide a list of abbreviations and accounts for sentences ended by terminators in wrappers, such as a period and quote: [.", ?', .)].

abbreviations = {'dr.': 'doctor', 'mr.': 'mister', 'bro.': 'brother', 'bro': 'brother', 'mrs.': 'mistress', 'ms.': 'miss', 'jr.': 'junior', 'sr.': 'senior', 'i.e.': 'for example', 'e.g.': 'for example', 'vs.': 'versus'}
terminators = ['.', '!', '?']
wrappers = ['"', "'", ')', ']', '}']


def find_sentences(paragraph):
   end = True
   sentences = []
   while end > -1:
       end = find_sentence_end(paragraph)
       if end > -1:
           sentences.append(paragraph[end:].strip())
           paragraph = paragraph[:end]
   sentences.append(paragraph)
   sentences.reverse()
   return sentences


def find_sentence_end(paragraph):
    [possible_endings, contraction_locations] = [[], []]
    contractions = abbreviations.keys()
    sentence_terminators = terminators + [terminator + wrapper for wrapper in wrappers for terminator in terminators]
    for sentence_terminator in sentence_terminators:
        t_indices = list(find_all(paragraph, sentence_terminator))
        possible_endings.extend(([] if not len(t_indices) else [[i, len(sentence_terminator)] for i in t_indices]))
    for contraction in contractions:
        c_indices = list(find_all(paragraph, contraction))
        contraction_locations.extend(([] if not len(c_indices) else [i + len(contraction) for i in c_indices]))
    possible_endings = [pe for pe in possible_endings if pe[0] + pe[1] not in contraction_locations]
    if len(paragraph) in [pe[0] + pe[1] for pe in possible_endings]:
        max_end_start = max([pe[0] for pe in possible_endings])
        possible_endings = [pe for pe in possible_endings if pe[0] != max_end_start]
    possible_endings = [pe[0] + pe[1] for pe in possible_endings if sum(pe) > len(paragraph) or (sum(pe) < len(paragraph) and paragraph[sum(pe)] == ' ')]
    end = (-1 if not len(possible_endings) else max(possible_endings))
    return end


def find_all(a_str, sub):
    start = 0
    while True:
        start = a_str.find(sub, start)
        if start == -1:
            return
        yield start
        start += len(sub)

I used Karl's find_all function from this entry: Find all occurrences of a substring in Python

回答8:

I'm not great at regular expressions, but a simpler version, "brute force" actually, of above is

sentence = re.compile("([\'\"][A-Z]|([A-Z][a-z]*\. )|[A-Z])(([a-z]*\.[a-z]*\.)|([A-Za-z0-9]*\.[A-Za-z0-9])|([A-Z][a-z]*\. [A-Za-z]*)|[^\.?]|[A-Za-z])*[\.?]")

which means start acceptable units are '[A-Z] or "[A-Z]
please note, most regular expressions are greedy so the order is very important when we do |(or). That's, why I have written i.e. regular expression first, then is come forms like Inc.

回答9:

My example is based on the example of Ali, adapted to Brazilian Portuguese. Thanks Ali.

ABREVIACOES = ['sra?s?', 'exm[ao]s?', 'ns?', 'nos?', 'doc', 'ac', 'publ', 'ex', 'lv', 'vlr?', 'vls?',
               'exmo(a)', 'ilmo(a)', 'av', 'of', 'min', 'livr?', 'co?ls?', 'univ', 'resp', 'cli', 'lb',
               'dra?s?', '[a-z]+r\(as?\)', 'ed', 'pa?g', 'cod', 'prof', 'op', 'plan', 'edf?', 'func', 'ch',
               'arts?', 'artigs?', 'artg', 'pars?', 'rel', 'tel', 'res', '[a-z]', 'vls?', 'gab', 'bel',
               'ilm[oa]', 'parc', 'proc', 'adv', 'vols?', 'cels?', 'pp', 'ex[ao]', 'eg', 'pl', 'ref',
               '[0-9]+', 'reg', 'f[ilí]s?', 'inc', 'par', 'alin', 'fts', 'publ?', 'ex', 'v. em', 'v.rev']

ABREVIACOES_RGX = re.compile(r'(?:{})\.\s*$'.format('|\s'.join(ABREVIACOES)), re.IGNORECASE)

        def sentencas(texto, min_len=5):
            # baseado em https://stackoverflow.com/questions/25735644/python-regex-for-splitting-text-into-sentences-sentence-tokenizing
            texto = re.sub(r'\s\s+', ' ', texto)
            EndPunctuation = re.compile(r'([\.\?\!]\s+)')
            # print(NonEndings)
            parts = EndPunctuation.split(texto)
            sentencas = []
            sentence = []
            for part in parts:
                txt_sent = ''.join(sentence)
                q_len = len(txt_sent)
                if len(part) and len(sentence) and q_len >= min_len and \
                        EndPunctuation.match(sentence[-1]) and \
                        not ABREVIACOES_RGX.search(txt_sent):
                    sentencas.append(txt_sent)
                    sentence = []

                if len(part):
                    sentence.append(part)
            if sentence:
                sentencas.append(''.join(sentence))
            return sentencas

Full code in: https://github.com/luizanisio/comparador_elastic

回答10:

If you want to break up sentences at 3 periods (not sure if this is what you want) you can use this regular expresion:

import re

text = """\
Mr. Smith bought cheapsite.com for 1.5 million dollars, i.e. he paid a lot for it. Did he mind? Adam Jones Jr. thinks he didn't. In any case, this isn't true... Well, with a probability of .9 it isn't.
"""
sentences = re.split(r'\.{3}', text)

for stuff in sentences:
     print(stuff)

来源：https://stackoverflow.com/questions/25735644/python-regex-for-splitting-text-into-sentences-sentence-tokenizing

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