Split string at nth occurrence of a given character

Question

Is there a Python-way to split a string after the nth occurrence of a given delimiter?

Given a string:

'20_231_myString_234' 

It should be split into (with the delimiter being '_', after its second occurrence):

['20_231', 'myString_234'] 

Or is the only way to accomplish this to count, split and join?

Answer 1

>>> n = 2 >>> groups = text.split('_') >>> '_'.join(groups[:n]), '_'.join(groups[n:]) ('20_231', 'myString_234') 

Seems like this is the most readable way, the alternative is regex)

Answer 2

Using re to get a regex of the form ^((?:[^_]*_){n-1}[^_]*)_(.*) where n is a variable:

n=2 s='20_231_myString_234' m=re.match(r'^((?:[^_]*_){%d}[^_]*)_(.*)' % (n-1), s) if m: print m.groups() 

or have a nice function:

import re def nthofchar(s, c, n):     regex=r'^((?:[^%c]*%c){%d}[^%c]*)%c(.*)' % (c,c,n-1,c,c)     l = ()     m = re.match(regex, s)     if m: l = m.groups()     return l  s='20_231_myString_234' print nthofchar(s, '_', 2) 

Or without regexes, using iterative find:

def nth_split(s, delim, n):      p, c = -1, 0     while c < n:           p = s.index(delim, p + 1)         c += 1     return s[:p], s[p + 1:]   s1, s2 = nth_split('20_231_myString_234', '_', 2) print s1, ":", s2