grep up to 3 spaces with \s

Question

According to the following tutorials

1. https://linuxize.com/post/regular-expressions-in-grep/

\s Match a space.

and

2. https://www.guru99.com/linux-regular-expressions.html

Some interval regular expressions are:

Expression Description

{n} Matches the preceding character appearing 'n' times exactly

{n,m} Matches the preceding character appearing 'n' times but not more than m

{n, } Matches the preceding character only when it appears 'n' times or more

Sample file

wolf@linux:~$ cat space.txt
0space
1 spaces
2  spaces
3   spaces
4    spaces
wolf@linux:~$ 

I just want to grep up to 3 spaces only, minimum 1 space, maximum 3 spaces Unfortunately, it doesn't really work as expected.

wolf@linux:~$ cat space.txt | grep -P '\s{1,3}'
1 spaces
2  spaces
3   spaces
4    spaces
wolf@linux:~$ 

wolf@linux:~$ cat space.txt | grep -P '\s{3}'
3   spaces
4    spaces
wolf@linux:~$ 

wolf@linux:~$ cat space.txt | grep -P '\s{3,3}'
3   spaces
4    spaces
wolf@linux:~$ 

wolf@linux:~$ cat space.txt | grep -P '\s{0,3}'
0space
1 spaces
2  spaces
3   spaces
4    spaces
wolf@linux:~$ 

Desired Output

wolf@linux:~$ cat space.txt | grep -P '\s{0,3}' <- need to fix it here
1 spaces
2  spaces
3   spaces
wolf@linux:~$ 

Answer 1

you need:

grep -P '\S\s{1,3}\S' infile

\s matches a whitespace-character, not only a space.
\S matches a non-whitespace-character

in your attempt, you are not limiting that before &after your matches should not be a whitespace.

---

to filter on space only and avoid using PCRE, you can do:

grep '[^ ] \{1,3\}[^ ]' infile

or to work on lines having leading/trailing 1~3spaces:

grep '\([^ ]\|^\) \{1,3\}\([^ ]\|$\)' infile

input data (cat -e infile):

0space$
1 spaces$
2  spaces$
3   spaces$
4    spaces$
   3spaces$
    4space$
3spaces   $
4spaces    $

output:

1 spaces$
2  spaces$
3   spaces$
   3spaces$
3spaces   $

Answer 2

If you want to match on a sequence of 1 to 3 whitespace characters not surrounded by whitespaces, that's where you'd use Perl look-around operators:

grep -P '(?<!\s)\s{1,3}(?!\s)'

It matches on:

         1
1234567890123456789
    a b  c   d    e
     ^ ^^ ^^^ 

With standard grep, you could achieve the same effect with:

grep -E '(^|[^[:space:]])[[:space:]]{1,3}([^[:space:]]|$)'

This time we match on the sequence of 1 to 3 whitespace characters and the non-whitespace on either side (or the start (^) or end ($) of the subject).

         1
1234567890123456789
   a b  c   d    e
^^^^ ^^^^

(with -o (a GNU extension), you'd find it doesn't report a b as a was already matched earlier; when searching for more matches, it starts at the next character after the last match).

Without -E, you get basic regexps that don't have alternation operators (though some grep implementations support \| for that as an extension), but standardly, you could still do:

grep -x '\(.*[^[:space:]]\)\{0,1\}[[:space:]]\{1,3\}\([^[:space:]].*\)\{0,1\}'

This time, the regexp matches the whole line including the 1 to 3 spaces and an optional (\{0,1\} the equivalent of ERE ?) leading part to it ending in a non-whitespace and an optional part following it that starts with a non-whitespace.

         1
1234567890123456789
   a b  c   d    e
^^^^^^^^^^^^^^^^^^

In any case, those would still return lines that contain a sequence of 4 or more whitespaces as long as they also contain a sequence of 1 to 3 whitespace not surrounded by whitespaces.

If the point was to exclude lines containing sequences of 4 or more whitespaces then it would just be:

grep -vE '[[:space:]]{4}'

Or if you still require at least one whitespace, or in other words that the line contains one or more sequences of whitespace characters all of which have at least one whitespace but no more than 3:

grep -vE -e '[[:space:]]{4}' -e '^[^[:space:]]*$'

That is return all the lines except the ones that contain a sequence of 4 whitespace and the ones that are made exclusively of non-whitespaces.

Or again with Perl's look around operators:

grep -P '^(?=.*\s)(?!.*\s{4})'

That is match the beginning of the line, provided that it's followed by any amount of characters and a whitespace and that it's not followed by any amount of characters and a sequence of 4 whitespace.

Though it would be more legible with sed or awk where you can do both a positive and negative match in the same invocation:

awk '/[[:space:]]/ && ! /[[:space:]]{4}/'

sed '/[[:space:]]/!d; /[[:space:]]\{4\}/d'

Answer 3

You can come up from the opposite. Exclude lines with more than 3 spaces in the substring.

grep -Ev '\s{4,}'

-v Invert the sense of matching, to select non-matching lines.
You can insert anchors as non-whitespace characters

grep -E '\S\s{1,3}\S'

Answer 4

$ grep -E '[[:space:]]' < file |
  grep -vE '[[:space:]]{4}'
1 spaces
2  spaces
3   spaces

• first filter all lines with atleast 1 space char.
• from these, filter out all lines with 4 or more space characters.
• what remains is lines comprising 1 to 3 space characters.