How to find the length of a leading sequence in a string?

Question

I'd like to count the number of leading spaces in a string.What's the most Pythonic way of doing this?

>>>F(' ' * 5 + 'a')
5

(update) Here are timings of several of the answers:

import timeit
>>> timeit.timeit("s.index(re.search(r'\S',s).group())", number=10000, setup="import re;s='     a'")
0.027384042739868164
>>> timeit.timeit("len([i for i in itertools.takewhile(str.isspace,s)])", number=10000, setup="import itertools;s='     a'")
0.025166034698486328
>>> timeit.timeit("next(idx for idx,val in enumerate(s) if val != ' ')", number=10000, setup="s='     a'")
0.028306961059570312
>>> timeit.timeit("F('     a')", number=10000, setup="def F(s): return len(s)-len(s.lstrip(' '))")
0.0051808357238769531

Answer 1

Using re module

>>> s
'     a'
>>> import re
>>> s.index(re.search(r'\S',s).group())
5

Using itertools

>>> import itertools
>>> len([i for i in itertools.takewhile(str.isspace,s)])
5

The brute force way

>>> def F(s):
...     for i in s:
...          if i!=' ':
...               return s.index(i)
... 
>>> F(s)
5

Answer 2

>>> F = lambda x:len(x)-len(x.lstrip(' '))
>>> F(' ' * 5 + 'a')
5

If you need to find length of leading whitespaces, you can just remove ' ' from lstrip, which is now limited to spaces only

Answer 3

def f(string):
    try:
        return [c == ' ' for c in string].index(False)
    except ValueError:
        return -1

looks for the first character that doesn't match ' '

Answer 4

Use enumerate to iterate through the string s. As soon as you hit a non ' ' item in s, print that index, which is what you are looking for.

>>> for i, val in enumerate(s):
...    if val != ' ':  
...       print i  
...       break
...
5

Answer 5

Through re module. and using len function.

>>> s = '     a'
>>> len(re.sub(r'\S.*', '',s))
5

This would remove all the chars from the first non-space character upto the last. Then the len function applies on the resultant string giving you the length of leading spaces.

OR

>>> s = '     a'
>>> len(re.match(r'\s*', s).group())
5
>>> s = 'a'
>>> len(re.match(r'\s*', s).group())
0

This would match only zero or more leading spaces.

OR

Use re.match function's end() attribute.

>>> s = '     a'
>>> re.match(r'\s*', s).end()
5
>>> re.match(r'\s*', 'd').end()
0

Thanks to Adam

Answer 6

Ooh one more that no one has posted yet.

def leading_spaces(s):
    return next(idx for idx,val in enumerate(s) if val != " ")

This is almost certainly the fastest way to do it

Answer 7

A simple map and then sum:

>>> sum(map(bool, takewhile(str.isspace, '    a  ')))
4

Answer 8

Code:

>>> i = 0
>>> while i < len(s):
...     if s[i] != ' ':
...       print i
...       break  
...     i += 1

5

Answer 9

Strip the string on the left side using the first character:

>>> s = 'aaaaabaca'
>>> len(s) - len(s.lstrip(s[0]))
5

If you want to make it work for empty strings, simply add a check:

>>> s = ''
>>> len(s) - len(s.lstrip(s[0])) if s else 0
0

Answer 10

sorry for not being explicit enough....initialize a counter...you just iterate through the string and set a condition to check if each variable in the list a whitespace and the counter increment by 1...

counter = 0

sstring = " " * 5 + "a"

for x in sstring:

    if x == " ":
        counter += 1
    else:
        continue

print(counter)

i am sincerely sorry i dont know why my script keeps showing in this lousy format..i am kind of new here