python re backreference repeated elements

Question

Let's say I have a string like this...

myStr = 'START1(stuff); II(morestuff); 8(lessstuff)'

...and I want to extract the string immediately before the parentheses, as well as the string within the parentheses: 1, stuff, II, morestuff, 8, lessstuff. I can achieve this using split(';'), etc., but I want to see if I can do it in one fell swoop with re.search(). I have tried...

test = re.search( r'START(?:([I0-9]+)\(([^)]+?)\)(?:; )?)*', myStr ).groups()

...or in a more readable format...

test = re.search( r'''
                  START         # This part begins each string
                  (?:           # non-capturing group
                    ([I0-9]+)   # capture label before parentheses
                    \(
                      ([^)]+?)  # any characters between the parentheses
                    \)
                    (?:; )?     # semicolon + space delimiter
                  )*
                  ''', myStr, re.VERBOSE ).groups()

...but I only get the last hit: ('8', 'lessstuff'). Is there a way to backreference multiple hits of the same part of the expression?

If you're going to do that, it is imperative that you learn about the re.VERBOSE flag first: docs.python.org/2/library/re.html#re.VERBOSE ;-) — thebjorn
– thebjorn, Commented May 6, 2016 at 17:25
@heemayl Just 1. I could have left START off for the purposes of this question. — reynoldsnlp
– reynoldsnlp, Commented May 6, 2016 at 17:26

anubhava · Accepted Answer · 2016-05-06 17:26:56Z

3

You can use this regex in findall to capture your text:

>>> myStr = 'START1(stuff); II(morestuff); 8(lessstuff)'
>>> print re.findall(r'(?:START)?(\w+)\(([^)]*)\)', myStr)
[('1', 'stuff'), ('II', 'morestuff'), ('8', 'lessstuff')]

RegEx Demo

answered May 6, 2016 at 17:26

anubhava

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1 Comment

reynoldsnlp Over a year ago

Thanks! I had forgotten about findall()!

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