Regular Expression WhiteSpace Character that Excludes Newline [duplicate] - regex

I sometimes want to match whitespace but not newline.
So far I've been resorting to [ \t]. Is there a less awkward way?

Use a double-negative:
/[^\S\r\n]/
That is, not-not-whitespace (the capital S complements) or not-carriage-return or not-newline. Distributing the outer not (i.e., the complementing ^ in the character class) with De Morgan's law, this is equivalent to “whitespace but not carriage return or newline.” Including both \r and \n in the pattern correctly handles all of Unix (LF), classic Mac OS (CR), and DOS-ish (CR LF) newline conventions.
No need to take my word for it:
#! /usr/bin/env perl
use strict;
use warnings;
use 5.005; # for qr//
my $ws_not_crlf = qr/[^\S\r\n]/;
for (' ', '\f', '\t', '\r', '\n') {
my $qq = qq["$_"];
printf "%-4s => %s\n", $qq,
(eval $qq) =~ $ws_not_crlf ? "match" : "no match";
}
Output:
" " => match
"\f" => match
"\t" => match
"\r" => no match
"\n" => no match
Note the exclusion of vertical tab, but this is addressed in v5.18.
Before objecting too harshly, the Perl documentation uses the same technique. A footnote in the “Whitespace” section of perlrecharclass reads
Prior to Perl v5.18, \s did not match the vertical tab. [^\S\cK] (obscurely) matches what \s traditionally did.
The same section of perlrecharclass also suggests other approaches that won’t offend language teachers’ opposition to double-negatives.
Outside locale and Unicode rules or when the /a switch is in effect, “\s matches [\t\n\f\r ] and, starting in Perl v5.18, the vertical tab, \cK.” Discard \r and \n to leave /[\t\f\cK ]/ for matching whitespace but not newline.
If your text is Unicode, use code similar to the sub below to construct a pattern from the table in the aforementioned documentation section.
sub ws_not_nl {
local($_) = <<'EOTable';
0x0009 CHARACTER TABULATION h s
0x000a LINE FEED (LF) vs
0x000b LINE TABULATION vs [1]
0x000c FORM FEED (FF) vs
0x000d CARRIAGE RETURN (CR) vs
0x0020 SPACE h s
0x0085 NEXT LINE (NEL) vs [2]
0x00a0 NO-BREAK SPACE h s [2]
0x1680 OGHAM SPACE MARK h s
0x2000 EN QUAD h s
0x2001 EM QUAD h s
0x2002 EN SPACE h s
0x2003 EM SPACE h s
0x2004 THREE-PER-EM SPACE h s
0x2005 FOUR-PER-EM SPACE h s
0x2006 SIX-PER-EM SPACE h s
0x2007 FIGURE SPACE h s
0x2008 PUNCTUATION SPACE h s
0x2009 THIN SPACE h s
0x200a HAIR SPACE h s
0x2028 LINE SEPARATOR vs
0x2029 PARAGRAPH SEPARATOR vs
0x202f NARROW NO-BREAK SPACE h s
0x205f MEDIUM MATHEMATICAL SPACE h s
0x3000 IDEOGRAPHIC SPACE h s
EOTable
my $class;
while (/^0x([0-9a-f]{4})\s+([A-Z\s]+)/mg) {
my($hex,$name) = ($1,$2);
next if $name =~ /\b(?:CR|NL|NEL|SEPARATOR)\b/;
$class .= "\\N{U+$hex}";
}
qr/[$class]/u;
}
Other Applications
The double-negative trick is also handy for matching alphabetic characters too. Remember that \w matches “word characters,” alphabetic characters and digits and underscore. We ugly-Americans sometimes want to write it as, say,
if (/[A-Za-z]+/) { ... }
but a double-negative character-class can respect the locale:
if (/[^\W\d_]+/) { ... }
Expressing “a word character but not digit or underscore” this way is a bit opaque. A POSIX character-class communicates the intent more directly
if (/[[:alpha:]]+/) { ... }
or with a Unicode property as szbalint suggested
if (/\p{Letter}+/) { ... }

Perl versions 5.10 and later support subsidiary vertical and horizontal character classes, \v and \h, as well as the generic whitespace character class \s
The cleanest solution is to use the horizontal whitespace character class \h. This will match tab and space from the ASCII set, non-breaking space from extended ASCII, or any of these Unicode characters
U+0009 CHARACTER TABULATION
U+0020 SPACE
U+00A0 NO-BREAK SPACE (not matched by \s)
U+1680 OGHAM SPACE MARK
U+2000 EN QUAD
U+2001 EM QUAD
U+2002 EN SPACE
U+2003 EM SPACE
U+2004 THREE-PER-EM SPACE
U+2005 FOUR-PER-EM SPACE
U+2006 SIX-PER-EM SPACE
U+2007 FIGURE SPACE
U+2008 PUNCTUATION SPACE
U+2009 THIN SPACE
U+200A HAIR SPACE
U+202F NARROW NO-BREAK SPACE
U+205F MEDIUM MATHEMATICAL SPACE
U+3000 IDEOGRAPHIC SPACE
The vertical space pattern \v is less useful, but matches these characters
U+000A LINE FEED
U+000B LINE TABULATION
U+000C FORM FEED
U+000D CARRIAGE RETURN
U+0085 NEXT LINE (not matched by \s)
U+2028 LINE SEPARATOR
U+2029 PARAGRAPH SEPARATOR
There are seven vertical whitespace characters which match \v and eighteen horizontal ones which match \h. \s matches twenty-three characters
All whitespace characters are either vertical or horizontal with no overlap, but they are not proper subsets because \h also matches U+00A0 NO-BREAK SPACE, and \v also matches U+0085 NEXT LINE, neither of which are matched by \s

A variation on Greg’s answer that includes carriage returns too:
/[^\S\r\n]/
This regex is safer than /[^\S\n]/ with no \r. My reasoning is that Windows uses \r\n for newlines, and Mac OS 9 used \r. You’re unlikely to find \r without \n nowadays, but if you do find it, it couldn’t mean anything but a newline. Thus, since \r can mean a newline, we should exclude it too.

The below regex would match white spaces but not of a new line character.
(?:(?!\n)\s)
DEMO
If you want to add carriage return also then add \r with the | operator inside the negative lookahead.
(?:(?![\n\r])\s)
DEMO
Add + after the non-capturing group to match one or more white spaces.
(?:(?![\n\r])\s)+
DEMO
I don't know why you people failed to mention the POSIX character class [[:blank:]] which matches any horizontal whitespaces (spaces and tabs). This POSIX chracter class would work on BRE(Basic REgular Expressions), ERE(Extended Regular Expression), PCRE(Perl Compatible Regular Expression).
DEMO

What you are looking for is the POSIX blank character class. In Perl it is referenced as:
[[:blank:]]
in Java (don't forget to enable UNICODE_CHARACTER_CLASS):
\p{Blank}
Compared to the similar \h, POSIX blank is supported by a few more regex engines (reference). A major benefit is that its definition is fixed in Annex C: Compatibility Properties of Unicode Regular Expressions and standard across all regex flavors that support Unicode. (In Perl, for example, \h chooses to additionally include the MONGOLIAN VOWEL SEPARATOR.) However, an argument in favor of \h is that it always detects Unicode characters (even if the engines don't agree on which), while POSIX character classes are often by default ASCII-only (as in Java).
But the problem is that even sticking to Unicode doesn't solve the issue 100%. Consider the following characters which are not considered whitespace in Unicode:
U+180E MONGOLIAN VOWEL SEPARATOR
U+200B ZERO WIDTH SPACE
U+200C ZERO WIDTH NON-JOINER
U+200D ZERO WIDTH JOINER
U+2060 WORD JOINER
U+FEFF ZERO WIDTH NON-BREAKING SPACE
Taken from https://en.wikipedia.org/wiki/White-space_character
The aforementioned Mongolian vowel separator isn't included for what is probably a good reason. It, along with 200C and 200D, occur within words (AFAIK), and therefore breaks the cardinal rule that all other whitespace obeys: you can tokenize with it. They're more like modifiers. However, ZERO WIDTH SPACE, WORD JOINER, and ZERO WIDTH NON-BREAKING SPACE (if it used as other than a byte-order mark) fit the whitespace rule in my book. Therefore, I include them in my horizontal whitespace character class.
In Java:
static public final String HORIZONTAL_WHITESPACE = "[\\p{Blank}\\u200B\\u2060\\uFFEF]"

Put the regex below in the find section and select Regular Expression from "Search Mode":
[^\S\r\n]+

m/ /g just give space in / /, and it will work. Or use \S — it will replace all the special characters like tab, newlines, spaces, and so on.

Related

Vim regex erasing everything between colon and a space or a newline

I'm trying to use vim regex to erase everything after a colon : and a space or newline character. Below is the text that I'm working with.
ablatives ablative:ablative_A s:+PL
abounded abound:abound_V ed:+PAST
abrogate abrogate:abrogate_V
abusing ab:ab_p us:use_V ing:+PCP1
accents' accent:accent_N s:+PL ':+GEN
accorded accord:accord_V ed:+PAST
So what I want to get from this is the following:
ablatives ablative s
abounded abound ed
abrogate abrogate
abusing ab us ing
accents' accent s '
accorded accord ed
I'm pretty lost on this one but I did trying the statement below:
:s/\:. / /g
I'm trying use that to get at least one of the patterns that I need.
Simply, you can do :
:%s/:\S*//g
\S non-whitespace character;
This seems to give the results you want:
:%s/:.\{-}\([ \n]\)/\1/g
: – literal :. I'm not sure why you escaped that in your question, since you don't need to.
. – mach any character.
\{-} – repeated zero or more times as little as possible ("ungreedy", like *? in many other regexp engines).
\( – start new subgroup for the reference in the replacement pattern.
[ – start "match any of these characters.
– literal space.
\n – newline.
] – end [.
\) – end subgroup.
In the replacement pattern we use \1 to insert either a space or newline, depending on what we replaced.
You can also use \_s instead of [ \n], this will match all spaces, tabs, and newlines. I personally prefer [ \n] since that's more portable across regexp engines (whereas \_s is a Vim-specific construct).
see :help <atom> for more information on any of the above.
:%s/:[^ ]*//g
This deletes : followed by zero or more non-space characters. Newline character at end of line won't be affected

What's the difference between [:space:] and [:blank:]?

From the A Brief Introduction to Regular Expressions
[:blank:] matches a space or a tab.
[:space:] matches whitespace characters (space and horizontal tab).
To me both definitions are the same and I was wondering if they are really duplicates?
If they are different, what are the differences?
For the GNU tools the following from grep.info applies:
[:blank:]
Blank characters: space and tab.
[:space:]
Space characters: in the 'C' locale, this is tab, newline,
vertical tab, form feed, carriage return, and space.
You can find the section with this command:
info grep 'Regular Expressions' 'Character Classes and Bracket Expressions'
A better explanation of what they each match is available here
http://www.regular-expressions.info/posixbrackets.html
The biggest difference is that [:space:] will also match items like new line characters
A space means to pressing a space bar and tab
A white space means it carries the newline,tab,form feed and carriage return and also space that's all.

Removing repeated characters, including spaces, in one line

I currently have a string, say $line='55.25040882, 3,,,,,,', that I want to remove all whitespace and repeated commas and periods from. Currently, I have:
$line =~ s/[.,]{2,}//;
$line =~ s/\s{1,}//;
Which works, as I get '55.25040882,3', but when I try
$line =~ s/[.,\s]{2,}//;
It pulls out the ", " and leaves the ",,,,,,". I want to retain the first comma and just get rid of the whitespace.
Is there a way to elegantly do this with one line of regex? Please let me know if I need to provide additional information.
EDIT: Since there were so many solutions, I decided to update my question with the answer below:
$line =~ s/([.,])\1{1,}| |\t//g;
This removes all repeated periods and commas, removes all spaces and tabs, while retaining the \r and \n characters. There are so many ways to do this, but this is the one I settled for. Thanks so much!
This is mostly a critique of Rohit's answer, which seems to contain several misconceptions about character class syntax, especially the negation operator (^). Specifically:
[(^\n^\r)\s] matches ( or ^ or ) or any whitespace character, including linefeed (\n) and carriage return (\r). In fact, they're each specified twice (since \s matches them too), though the class still only consumes one character at a time.
^[\n\r]|\s matches a linefeed or carriage return at the beginning of the string, or any whitespace character anywhere (which makes the first part redundant, since any whitespace character includes linefeed and carriage return, and anywhere includes the beginning of the string).
Inside a character class, the caret (^) negates the meaning of everything that follows iff it appears immediately after the opening [; anywhere else, it's just a caret. All other metacharacters except \ lose their special meanings entirely inside character classes. (But the normally non-special characters, - and ], become special.)
Outside a character class, ^ is an anchor.
Here's how I would write the regex:
$line =~ s/([.,])\1+|\h+//g;
Explanation:
Since you finally went with ([.,])\1{1,}, I assume you want to match repeated periods or repeated commas, not things like ., or ,.. Success with regexes means learning to look at text the way the regex engine does, and it's not intuitive. You'll help yourself a lot if you try to describe each problem the way the regex engine would, if it could speak.
{1,} is not incorrect, but why add all that clutter to your regex when + does the same thing?
\h matches horizontal whitespace, which includes spaces and tabs, but not linefeeds or carriage returns. (That only works in Perl, AFAIK. In Ruby/Oniguruma, \h matches a hex digit; in every other flavor I know of, it's a syntax error.)
You can try using: -
my $line='55.25040...882, 3,,,,,,';
$line =~ s/[^\S\n\r]|[.,]{2,}//g; # Negates non-whitespace char, \n and \r
print $line
OUTPUT: -
55.25040882,3
[^\S\n\r]|[.,]{2,} -> This means either [^\S\n\r] or [.,]{2,}
[.,]{2,} -> This means replace , or . if there is more than 2 in the same
line.
[^\S\n\r] -> Means negate all whitespace character, linefeed, and newline.

Vim regex not matching spaces in a character class

I'm using vim to do a search and replace with this command:
%s/lambda\s*{\([\n\s\S]\)*//gc
I'm trying to match for all word, endline and whitespace characters after a {. For instance, the entirety of this line should match:
lambda {
FactoryGirl.create ...
Instead, it only matches up to the newline and no spaces before FactoryGirl. I've tried manually replacing all the spaces before, just in case there were tab characters instead, but no dice. Can anyone explain why this doesn't work?
The \s is an atom for whitespace; \n, though it looks similar, syntactically is an escape sequence for a newline character. Inside the collection atom [...], you cannot include other atoms, only characters (including some special ones like \n. From :help /[]:
The following translations are accepted when the 'l' flag is not
included in 'cpoptions' {not in Vi}:
\e <Esc>
\t <Tab>
\r <CR> (NOT end-of-line!)
\b <BS>
\n line break, see above |/[\n]|
\d123 decimal number of character
\o40 octal number of character up to 0377
\x20 hexadecimal number of character up to 0xff
\u20AC hex. number of multibyte character up to 0xffff
\U1234 hex. number of multibyte character up to 0xffffffff
NOTE: The other backslash codes mentioned above do not work inside
[]!
So, either specify the whitespace characters literally [ \t\n...], use the corresponding character class expression [[:space:]...], or combine the atom with the collection via logical or \%(\s\|[...]\).
Vim interprets characters inside of the [ ... ] character classes differently. It's not literally, since that regex wouldn't fully match lambda {sss or lambda {\\\. What \s and \S are interpreted as...I still can't explain.
However, I was able to achieve nearly what I wanted with:
%s/lambda\s*{\([\n a-zA-z]\)*//gc
That ignores punctuation, which I wanted. This works, but is dangerous:
%s/lambda\s*{\([\n a-zA-z]\|.\)*//gc
Because adding on a character after the last character like } causes vim to hang while globbing. So my solution was to add the punctuation I needed into the character class.

regular expression confusion \s and " "

In regular expression, i know when use \s to represent a space, but, in following case, would they be different:
/a\sb/ ---with a \s
/a b/ ---with empty field
thanks a lot if you can explain to me.
The \s character class matches all "whitespace characters," not just spaces. This includes tabs (\t), and if multiline matching is allowed, it includes carriage return (\r) and newline (\n). Theoretically, if your regular expression engine handles unicode, there are also unicode whitespace characters that \s can match, though your mileage may vary.
So with a string like "a\t b", you can match it with the regex /a\s+b/, in case that is useful to you.