how to match only once occurrence of a double space of a line? - python-3.x

Question

line A
foo bar bar foo bar foo
line B
foo bar bar foo
In line A, there are multiple occurrence of double space.
I only want to match lines like line B which has only once double space occurrence.
I tried
^.*\s{2}.*$
but it will match both.
How may I have the desired output? Thank you.

Answer 1

If you wish to match strings that contain no more than one string of two or more spaces between words you could use following regular expression.
r'^(?!(?:.*(?<! ) {2,}(?! )){2})'
Start your engine!
Note that this expression matches
abc de fgh
where there are four spaces between 'c' and 'd'.
Python's regex engine performs the following operations.
^
(?! : begin negative lookahead
(?: : begin non-capture group
.* : match 0+ characters other than line terminators
(?<! : begin negative lookbehind
[ ]{2,} : match 2+ spaces
(?! ) : negative lookahead asserts match is not followed by a space
) : end negative lookbehind
) : end non-capture group
{2} : execute non-capture group twice
) : end negative lookahead

Answer 2

You can do:
^(?!.*[ \t]{2,}.*[ \t]{2,})
# Negative look ahead assertion that states 'only start the match
# on this line IF there are NOT 2 (or potentially more) breaks with
# two (or potentially more) of tabs or spaces'.
Demo 1
If you want to require ONE double space in the line but not more:
^(?=.*[ \t]{2,})(?!.*[ \t]{2,}.*[ \t]{2,})
# Positive look ahead that states 'only start this match if there is
# at least one break with two tabs or spaces'
# BUT
# Negative look ahead assertion that states 'only start the match
# on this line IF there are NOT 2 (or potentially more) breaks with
# two (or potentially more) of tabs or spaces'.
Demo 2
If you want to limit to only two spaces (not tabs and not more than 2 spaces):
^(?=.*[ ]{2})(?!.*[ ]{2}.*[ ]{2})
# Same as above but remove the tabs as part of the assertion
Demo 3
Note: In your regex you have \s as the class for a space. That also matches [\r\n\t\f\v ] so both horizontal and vertical space characters.
Note 2:
You can do this without a regex as well (assuming you only want lines that have 1 and only 1 double space in them):
txt='''\
line A
foo bar bar foo bar foo
line B
foo bar bar foo'''
>>> [line for line in txt.splitlines() if len(line.split(' '))==2]
['foo bar bar foo']

Answer 3

You can get the match without lookarounds by starting the match with 1+ non whitespace chars.
Then optionally repeat a single whitespace char followed by non whitespace chars before and after matching a double whitespace char.
The negated character class [^\S\r\n] will match any whitespace chars except a newline or carriage return. If you want to allow matching newlines as well, you could use \s
^\S+(?:[^\S\r\n]\S+)*[^\S\r\n]{2}(?:\S+[^\S\r\n])*\S+$
Explanation
^ Start of string
\S+ Match 1+ non whitespace chars
(?: Non capture group
[^\S\r\n]\S+ Match a whitespace char without a newline
)* Close group and repeat 0+ times
[^\S\r\n]{2} Match the 2 whitespace chars without a newline
(?: Non capture group
\S+[^\S\r\n] Match 1+ non whitespace chars followed by a whitespace char without a newline
)* Close group a and repeat 1+ times
\S+ Match 1+ non whitespace chars
$ End of string
Regex demo