Question

I have a string in the format:

t='@abc @def Hello this part is text' 

I want to get this:

l=['abc', 'def']  s='Hello this part is text' 

I did this:

a=t[t.find(' ',t.rfind('@')):].strip() s=t[:t.find(' ',t.rfind('@'))].strip() b=a.split('@') l=[i.strip() for i in b][1:] 

It works for the most part, but it fails when the text part has the ‘@’. Eg, when:

t='@abc @def My email is [e-mail]' 

it fails. The @names are there in the beginning and there can be text after @names, which may possibly contain @.

Clearly I can append initally with a space and find out first word without ‘@’. But that doesn’t seem an elegant solution.

What is a pythonic way of solving this?

Answer 1

Building unashamedly on MrTopf’s effort:

import re rx = re.compile('((?:@\w+ +)+)(.*)') t='@abc   @def  @xyz Hello this part is text and my email is foo@ba.r' a,s = rx.match(t).groups() l = re.split('[@ ]+',a)[1:-1] print l print s 

prints:

[‘abc’, ‘def’, ‘xyz’]
Hello this part is text and my email is foo@ba.r

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Justly called to account by hasen j, let me clarify how this works:

/@\w+ +/ 

matches a single tag – @ followed by at least one alphanumeric or _ followed by at least one space character. + is greedy, so if there is more than one space, it will grab them all.

To match any number of these tags, we need to add a plus (one or more things) to the pattern for tag; so we need to group it with parentheses:

/(@\w+ +)+/ 

which matches one-or-more tags, and, being greedy, matches all of them. However, those parentheses now fiddle around with our capture groups, so we undo that by making them into an anonymous group:

/(?:@\w+ +)+/ 

Finally, we make that into a capture group and add another to sweep up the rest:

/((?:@\w+ +)+)(.*)/ 

A last breakdown to sum up:

((?:@\w+ +)+)(.*)  (?:@\w+ +)+  (  @\w+ +)     @\w+ + 

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Note that in reviewing this, I’ve improved it – \w didn’t need to be in a set, and it now allows for multiple spaces between tags. Thanks, hasen-j!