基于空白和尾随标点符号化？

Question

I'm trying to come up with the regular expression to split a string up into a list based on white space or trailing punctuation.

eg

s = 'hel-lo  this has whi(.)te, space. very \n good'

What I want is

['hel-lo', 'this', 'has', 'whi(.)te', ',', 'space', '.', 'very', 'good']

s.split() gets me most of the way there, except it doesn't take care of the trailing whitespace.

Answer 1

import re
s = 'hel-lo  this has whi(.)te, space. very \n good'
[x for x in re.split(r"([.,!?]+)?\s+", s) if x]
# => ['hel-lo', 'this', 'has', 'whi(.)te', ',', 'space', '.', 'very', 'good']

You might need to tweak what "punctuation" is.

Answer 2

Rough solution using spacy . It works pretty good with tokenizing word already.

import spacy
s = 'hel-lo  this has whi(.)te, space. very \n good'
nlp = spacy.load('en') 
ls = [t.text for t in nlp(s) if t.text.strip()]

>> ['hel', '-', 'lo', 'this', 'has', 'whi(.)te', ',', 'space', '.', 'very', 'good']

However, it also tokenize words between - so I borrow solution from here to merge words between - back together.

merge = [(i-1, i+2) for i, s in enumerate(ls) if i >= 1 and s == '-']
for t in merge[::-1]:
    merged = ''.join(ls[t[0]:t[1]])
    ls[t[0]:t[1]] = [merged]

>> ['hel-lo', 'this', 'has', 'whi(.)te', ',', 'space', '.', 'very', 'good']

Answer 3

I am using Python 3.6.1.

import re

s = 'hel-lo  this has whi(.)te, space. very \n good'
a = [] # this list stores the items
for i in s.split(): # split on whitespaces
    j = re.split('(\,|\.)$',i) # split on your definition of trailing punctuation marks
    if len(j) > 1:
        a.extend(j[:-1])
    else:
        a.append(i)
 # a -> ['hel-lo', 'this', 'has', 'whi(.)te', ',', 'space', '.', 'very', 'good']