How can I easily check if a string starts with 4N spaces?

Question

How can I easily check if a string starts with 4*N spaces, where N is a positive integer?

My current code is:

def StartsWith4Nspaces(string):
    count = 0
    for c in string:
        if c == ' ':
            count += 1
        else:
            break
    return count > 0 and count % 4 == 0

Is there a more Pythonic way to write this down?

I'm kinda hoping for a single statement (though anything cleaner than the above would be great).

Thank you.

Answer 1

You can just check for it like so:

my_string[:4*N] == ' ' * 4*N

You can also convert this check of yours into a lambda:

check = lambda my_string, N: my_string[:4*N] == ' ' * 4*N

and call it as:

check('  asdas', 2) # -> True
check('  asdas', 3) # -> False

Or if you want to hardcode the N for whatever reason (N = 3):

check = lambda my_string: my_string[:12] == ' ' * 12

---

EDIT: If the 4Nth + 1 character is required to not be a space, you can incorporate that into your lambda:

check_strict = lambda my_string, N: my_string[:4*N] == ' ' * 4*N and my_string[4*N + 1] != ' '

or

check_strict = lambda my_string: my_string[:12] == ' ' * 12 and my_string[13] != ' '

Answer 2

Using a regular expression works decently for this:

>>> re.match('(?: {4})*(?! )', '')
<_sre.SRE_Match object at 0x7fef988e4780>
>>> re.match('(?: {4})*(?! )', '  ')
>>> re.match('(?: {4})*(?! )', '    ')
<_sre.SRE_Match object at 0x7fef988e4718>
>>> re.match('(?: {4})*(?! )', 'foo')
<_sre.SRE_Match object at 0x7fef988e4780>
>>> re.match('(?: {4})*(?! )', '  foo')
>>> re.match('(?: {4})*(?! )', '    foo')
<_sre.SRE_Match object at 0x7fef988e4718>
>>> re.match('(?: {4})*(?! )', '      foo')
>>> re.match('(?: {4})*(?! )', '        foo')
<_sre.SRE_Match object at 0x7fef988e4780>

Note that this will allow N to be 0, and works with strings that only contain spaces. A valid match is considered true, but you can pass the result to bool() if you want it to be strictly a bool. Replacing the * with a + will force N to be greater than 0.

Answer 3

You can use the lstrip method to strip starting whitespace and then compare the lengths of the stripped and original strings:

s = string.lstrip()
return ((len(string) - len(s)) % 4 == 0 and (len(string) - len(s) != 0)

(You could even make it one line by not setting a variable for s.)

Answer 4

def startsWith4Nspaces(s):
    if not s: return False

    numLeadingSpaces = len(s) - len(s.lstrip(' '))
    if not numLeadingSpaces: return False
    if numLeadingSpaces%4: return False
    return True