Dividing a line with dots into words, but with a special case

Question

Not sure if there is an easy way to break the following line:

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

In it:

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

Note. I want to keep 'classes[cost=15.00]' intact.

Answer 1

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

More specific version:

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

Verbose:

 >>> 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

Skip dots in brackets:

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

Output:

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

Answer 3

This can become messy in a hurry, you may need to actually parse this line, and not just split it:

 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, ie, 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) 

Print

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

(It's easy to put "classes [cost = 15.00]" back together if you need to.)