Lets Say we have Zaptoit:685158:[email protected]
How do you split so it only be left 685158:[email protected]
+3
A:
Another solution:
s = 'Zaptoit:685158:[email protected]'
s.split(':', 1)[1]
Federico Ramponi
2009-01-12 19:15:09
+7
A:
>>> s = 'Zaptoit:685158:[email protected]'
>>> s.split( ':', 1 )[1]
'685158:[email protected]'
Graeme Perrow
2009-01-12 19:15:09
Note that it's not really good practice to use the variable name "str" since str() is a builtin.
Jay
2009-01-12 22:31:55
Thanks Jay - I've updated the code.
Graeme Perrow
2009-01-13 14:52:39
@PEZ: better being stopping the match on the first ':' instead of the last one
orip
2009-01-12 23:22:53
@orip: I think you are mistaken -- the question mark makes it a non-greedy match that will stop at the first colon, as intended
scrible
2009-01-13 04:12:30
It doesn't matter in this case, but in general, if you have choice between a negated character class and a reluctant quantifier (eg, '.*?'), the char class tends to be faster, less memory-intensive, and most importantly, more predictable.
Alan Moore
2009-01-13 15:46:18
To me it's about clarity. Clearer is better. In this case .*?: is perfectly predictable and very clear. "Everything up to the first colon" is clearer than "Everything that's not a colon up to the first colon".
PEZ
2009-01-13 16:16:01
+3
A:
Another method, without using split:
s = 'Zaptoit:685158:[email protected]'
s[s.find(':')+1:]
Ex:
>>> s = 'Zaptoit:685158:[email protected]'
>>> s[s.find(':')+1:]
'685158:[email protected]'
Jay
2009-01-12 19:59:47