Split string at nth occurrence of a given character

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?

>>> 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)

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

I like this solution because it works without any actuall regex and can easiely be adapted to another "nth" or delimiter.

import re

string = "20_231_myString_234"
occur = 2  # on which occourence you want to split

indices = [x.start() for x in re.finditer("_", string)]
part1 = string[0:indices[occur-1]]
part2 = string[indices[occur-1]+1:]

print (part1, ' ', part2)