Returning the lowest index for the first non whitespace character in a string in Python

Question

What's the shortest way to do this in Python?

string = "   xyz"

must return index = 3

Answer 1

>>> next(i for i, j in enumerate('   xyz') if j.strip())
3

or

>>> next(i for i, j in enumerate('   xyz') if j not in string.whitespace)
3

in versions of Python < 2.5 you'll have to do:

(...).next()

Answer 2

>>> s = "   xyz"
>>> len(s) - len(s.lstrip())
3

Answer 3

>>> string = "   xyz"
>>> next(idx for idx, chr in enumerate(string) if not chr.isspace())
3

Answer 4

>>> string = "   xyz"
>>> map(str.isspace,string).index(False)
3

Answer 5

Looks like the "regexes can do anything" brigade have taken the day off, so I'll fill in:

>>> tests = [u'foo', u' foo', u'\xA0foo']
>>> import re
>>> for test in tests:
...     print len(re.match(r"\s*", test, re.UNICODE).group(0))
...
0
1
1
>>>

FWIW: time taken is O(the_answer), not O(len(input_string))

Answer 6

import re
def prefix_length(s):
   m = re.match('(\s+)', s)
   if m:
      return len(m.group(0))
   return 0