使用正则表达式进行句子分割

Question

我有一些文本（SMS）消息，我想使用句点（'.'）作为分隔符对它们进行分段。我无法处理以下类型的消息。如何使用 Python 中的正则表达式对这些消息进行分段。

分割前：

'hyper count 16.8mmol/l.plz review b4 5pm.just to inform u.thank u'
'no of beds 8.please inform person in-charge.tq'

分割后：

'hyper count 16.8mmol/l' 'plz review b4 5pm' 'just to inform u' 'thank u'
'no of beds 8' 'please inform person in-charge' 'tq'

每行都是一条单独的消息

更新：

我正在做自然语言处理，我觉得可以处理'16.8mmmol/l'和'床数8.2杯茶。' 同样。 80% 的准确率对我来说已经足够了，但我想尽可能减少误报。

Answer 1

几周前，我搜索了一个正则表达式，它可以捕获字符串中代表数字的每个字符串，无论数字的书写形式如何，甚至是科学计数法中的数字，甚至是带有逗号的印度数字：请参阅 此线程

我在下面的代码中使用这个正则表达式来解决您的问题。

与其他答案相反，在我的解决方案中， '8.' 中的点不被视为必须进行分割的点，因为它可以被读取为后面没有数字的浮点数点。

import re

regx = re.compile('(?<![\d.])(?!\.\.)'
                  '(?<![\d.][eE][+-])(?<![\d.][eE])(?<!\d[.,])'
                  '' #---------------------------------
                  '([+-]?)'
                  '(?![\d,]*?\.[\d,]*?\.[\d,]*?)'
                  '(?:0|,(?=0)|(?<!\d),)*'
                  '(?:'
                  '((?:\d(?!\.[1-9])|,(?=\d))+)[.,]?'
                  '|\.(0)'
                  '|((?<!\.)\.\d+?)'
                  '|([\d,]+\.\d+?))'
                  '0*'
                  '' #---------------------------------
                  '(?:'
                  '([eE][+-]?)(?:0|,(?=0))*'
                  '(?:'
                  '(?!0+(?=\D|\Z))((?:\d(?!\.[1-9])|,(?=\d))+)[.,]?'
                  '|((?<!\.)\.(?!0+(?=\D|\Z))\d+?)'
                  '|([\d,]+\.(?!0+(?=\D|\Z))\d+?))'
                  '0*'
                  ')?'
                  '' #---------------------------------
                  '(?![.,]?\d)')



simpler_regex = re.compile('(?<![\d.])0*(?:'
                           '(\d+)\.?|\.(0)'
                           '|(\.\d+?)|(\d+\.\d+?)'
                           ')0*(?![\d.])')


def split_outnumb(string, regx=regx, a=0):
    excluded_pos = [x for mat in regx.finditer(string) for x in range(*mat.span()) if string[x]=='.']
    li = []
    for xdot in (x for x,c in enumerate(string) if c=='.' and x not in excluded_pos):
        li.append(string[a:xdot])
        a = xdot + 1
    li.append(string[a:])
    return li





for sentence in ('hyper count 16.8mmol/l.plz review b4 5pm.just to inform u.thank u',
                 'no of beds 8.please inform person in-charge.tq',
                 'no of beds 8.2 cups of tea.tarabada',
                 'this number .977 is a float',
                 'numbers 214.21E+45 , 478945.E-201 and .12478E+02 are in scientific.notation',
                 'an indian number 12,45,782.258 in this.sentence and 45,78,325. is another',
                 'no dot in this sentence',
                 ''):
    print 'sentence         =',sentence
    print 'splitted eyquem  =',split_outnumb(sentence)
    print 'splitted eyqu 2  =',split_outnumb(sentence,regx=simpler_regex)
    print 'splitted gurney  =',re.split(r"\.(?!\d)", sentence)
    print 'splitted stema   =',re.split('(?<!\d)\.|\.(?!\d)',sentence)
    print

结果

sentence         = hyper count 16.8mmol/l.plz review b4 5pm.just to inform u.thank u
splitted eyquem  = ['hyper count 16.8mmol/l', 'plz review b4 5pm', 'just to inform u', 'thank u']
splitted eyqu 2  = ['hyper count 16.8mmol/l', 'plz review b4 5pm', 'just to inform u', 'thank u']
splitted gurney  = ['hyper count 16.8mmol/l', 'plz review b4 5pm', 'just to inform u', 'thank u']
splitted stema   = ['hyper count 16.8mmol/l', 'plz review b4 5pm', 'just to inform u', 'thank u']

sentence         = no of beds 8.please inform person in-charge.tq
splitted eyquem  = ['no of beds 8.please inform person in-charge', 'tq']
splitted eyqu 2  = ['no of beds 8.please inform person in-charge', 'tq']
splitted gurney  = ['no of beds 8', 'please inform person in-charge', 'tq']
splitted stema   = ['no of beds 8', 'please inform person in-charge', 'tq']

sentence         = no of beds 8.2 cups of tea.tarabada
splitted eyquem  = ['no of beds 8.2 cups of tea', 'tarabada']
splitted eyqu 2  = ['no of beds 8.2 cups of tea', 'tarabada']
splitted gurney  = ['no of beds 8.2 cups of tea', 'tarabada']
splitted stema   = ['no of beds 8.2 cups of tea', 'tarabada']

sentence         = this number .977 is a float
splitted eyquem  = ['this number .977 is a float']
splitted eyqu 2  = ['this number .977 is a float']
splitted gurney  = ['this number .977 is a float']
splitted stema   = ['this number ', '977 is a float']

sentence         = numbers 214.21E+45 , 478945.E-201 and .12478E+02 are in scientific.notation
splitted eyquem  = ['numbers 214.21E+45 , 478945.E-201 and .12478E+02 are in scientific', 'notation']
splitted eyqu 2  = ['numbers 214.21E+45 , 478945.E-201 and .12478E+02 are in scientific', 'notation']
splitted gurney  = ['numbers 214.21E+45 , 478945', 'E-201 and .12478E+02 are in scientific', 'notation']
splitted stema   = ['numbers 214.21E+45 , 478945', 'E-201 and ', '12478E+02 are in scientific', 'notation']

sentence         = an indian number 12,45,782.258 in this.sentence and 45,78,325. is another
splitted eyquem  = ['an indian number 12,45,782.258 in this', 'sentence and 45,78,325. is another']
splitted eyqu 2  = ['an indian number 12,45,782.258 in this', 'sentence and 45,78,325. is another']
splitted gurney  = ['an indian number 12,45,782.258 in this', 'sentence and 45,78,325', ' is another']
splitted stema   = ['an indian number 12,45,782.258 in this', 'sentence and 45,78,325', ' is another']

sentence         = no dot in this sentence
splitted eyquem  = ['no dot in this sentence']
splitted eyqu 2  = ['no dot in this sentence']
splitted gurney  = ['no dot in this sentence']
splitted stema   = ['no dot in this sentence']

sentence         = 
splitted eyquem  = ['']
splitted eyqu 2  = ['']
splitted gurney  = ['']
splitted stema   = ['']

编辑1

我添加了一个 simler_regex 检测数字，来自我在 这个帖子

我没有检测到印度数字和科学计数法中的数字，但它实际上给出了相同的结果

Answer 2

您可以使用否定先行断言来匹配“。”后面不跟数字，然后使用 re.split ：

>>> import re
>>> splitter = r"\.(?!\d)"
>>> s = 'hyper count 16.8mmol/l.plz review b4 5pm.just to inform u.thank u'
>>> re.split(splitter, s)
['hyper count 16.8mmol/l', 'plz review b4 5pm', 'just to inform u', 'thank u']
>>> s = 'no of beds 8.please inform person in-charge.tq'
>>> re.split(splitter, s)
['no of beds 8', 'please inform person in-charge', 'tq']

Answer 3

环顾

re.split('(?<!\d)\.|\.(?!\d)', 'hyper count 16.8mmol/l.plz review b4 5pm.just to inform u.thank u')

四周确保一侧或另一侧都不是数字。因此这也涵盖了 16.8 情况。如果两边都有数字，这个表达式不会分裂。

Answer 4

这取决于你的具体句子，但你可以尝试：

.*?[a-zA-Z0-9]\.(?!\d)

看看是否有效。这将保留在引号中，但如果需要，您可以将其删除。

Answer 5

"...".split(".")

split 是一个 Python 内置函数，用于在特定字符处分隔字符串。