将点分隔的字符串拆分为单词，但有一个特殊情况

Question

不确定是否有一种简单的方法来拆分以下字符串：

'school.department.classes[cost=15.00].name'

进入这个：

['school', 'department', 'classes[cost=15.00]', 'name']

注意：我想保持'classes[cost=15.00]'不变。

Answer 1

>>> import re
>>> text = 'school.department.classes[cost=15.00].name'
>>> re.split(r'\.(?!\d)', text)
['school', 'department', 'classes[cost=15.00]', 'name']

更具体的版本：

>>> re.findall(r'([^.\[]+(?:\[[^\]]+\])?)(?:\.|$)', text)
['school', 'department', 'classes[cost=15.00]', 'name']

详细：

>>> re.findall(r'''(                      # main group
                    [^  .  \[    ]+       # 1 or more of anything except . or [
                    (?:                   # (non-capture) opitional [x=y,...]
                       \[                 # start [
                       [^   \]   ]+       # 1 or more of any non ]
                       \]                 # end ]
                    )?                    # this group [x=y,...] is optional
                   )                      # end main group
                   (?:\.|$)               # find a dot or the end of string
                ''', text, flags=re.VERBOSE)
['school', 'department', 'classes[cost=15.00]', 'name']

Answer 2

括号内的跳过点：

import re
s='school.department.classes[cost=15.00].name'
print re.split(r'[.](?![^][]*\])', s)

输出：

['school', 'department', 'classes[cost=15.00]', 'name']

Answer 3

这可能会很匆忙，您可能需要实际解析此字符串而不是仅仅将其拆分：

from pyparsing import (Forward,Suppress,Word,alphas,quotedString,
                        alphanums,Regex,oneOf,Group,delimitedList)


# define some basic punctuation, numerics, operators
LBRACK,RBRACK = map(Suppress, '[]')
ident = Word(alphas+'_',alphanums+'_')
real = Regex(r'[+-]?\d+\.\d*').setParseAction(lambda t:float(t[0]))
integer = Regex(r'[+-]?\d+').setParseAction(lambda t:int(t[0]))
compOper = oneOf('= != < > <= >=')

# a full reference may be composed of full references, i.e., a recursive
# grammar - forward declare a full reference
fullRef = Forward()

# a value in a filtering expression could be a full ref or numeric literal
value = fullRef | real | integer | quotedString
filterExpr = Group(value + compOper + value)

# a single dotted ref could be one with a bracketed filter expression
# (which we would want to keep together in a group) or just a plain identifier
ref = Group(ident + LBRACK + filterExpr + RBRACK) | ident

# now insert the definition of a fullRef, using '<<' instead of '='
fullRef << delimitedList(ref, '.')

# try it out
s = 'school.department.classes[cost=15.00].name'
print fullRef.parseString(s)
s = 'school[size > 10000].department[school.type="TECHNICAL"].classes[cost=15.00].name'
print fullRef.parseString(s)

打印：

['school', 'department', ['classes', ['cost', '=', 15.0]], 'name']
[['school', ['size', '>', 10000]], ['department', ['school', 'type', '=', '"TECHNICAL"']], ['classes', ['cost', '=', 15.0]], 'name']

（如果需要，将“课程[费用= 15.00]”重新组合起来并不困难。）

Answer 4

#最简单的句子拆分方法是使用.split('.') 如下所示：

s = 'school.department.classes[cost=15.00].name'

s.split('.')

这是您预期的 output：

['school', 'department', 'classes[cost=15', '00]', 'name']