Perl - Removing all special characters except a few - regex

So i came across a Perl regex "term" which allows you to remove all punctuation. Here is the code:
$string =~ s/[[:punct:]]//g;.
However this proceeds to remove all special characters. Is there a way that particular regex expression can be modified so that for example, it removes all special characters except hyphens. As i stated on my previous question with Perl, i am new to the language, thus obvious things don't come obvious to me. Thanks for all the help :_

Change your code like below to remove all the punctuations except hyphen,
$string =~ s/(?!-)[[:punct:]]//g;
DEMO
use strict;
use warnings;
my $string = "foo;\"-bar'.,...*(){}[]----";
$string =~ s/(?!-)[[:punct:]]//g;
print "$string\n";
Output:
foo-bar----

You may also use unicode property:
$string =~ s/[^-\PP]+//g;

Related

How do I search and replace with a regex and preserve blank spaces in Perl?

I'm using this code:
$text =~ s/\s(\w)/\u$1/g;
But This is an example
Become ThisIsAnExample
Instead of This Is An Example.
How to preserve blank spaces?
Use lookbehind.
$text =~ s/(?<!\S)(\w)/\u$1/g;
Or use the more efficient \K (Perl 5.10+).
$text =~ s/(?:^|\s)\K(\w)/\u$1/g;
Both of the solutions will make sure the first word is capitalized too. If that's not an issue, the second solution can be simplified to the following:
$text =~ s/\s\K(\w)/\u$1/g;
The matching contains the whitespace, the replacement doesn't.
$text =~ s/(\s)(\w)/$1\u$2/g;
Since \s contains different types of whitespace characters, if you want to keep it in your replacement, you need to capture it and put it back.
An alternative is to use word boundaries and full "words".
$text =~ s/\b(\w+)\b/\u$1/g;

How to substitute whitesapaces and tabs in a string with _ in perl?

$string = I am a boy
How to substitute whitespaces between words with underscore ?
You need a regular expression and the substitution operator to do that.
my $string = 'I am a boy';
$string =~ s/\s/_/g;
You can learn more about regex in perlre and perlretut. A nice tool to play around with is Rubular.
Also, your code will not compile. You need to quote your string, and you need to put a semicolon at the end.
$string = 'I am a boy';
$string =~ s/ /_/g;
$string =~ tr( \t)(_); # Double underscore not necessary as per Dave's comment
This is just to show another option in perl. I think Miguel Prz and imbabque showed more smarter ways, personally i follow the way imbabque showed.
my $str = "This is a test string";
$str =~ s/\p{Space}/_/g;
print $str."\n";
and the output is
This_is_a_test_string

Perl: How to replace only matched part of string?

I have a string foo_bar_not_needed_string_part_123. Now in this string I want to remove not_needed_string_part only when foo_ is followed by bar.
I used the below regex:
my $str = "foo_bar_not_needed_string_part_123";
say $str if $str =~ s/foo_(?=bar)bar_(.*?)_\d+//;
But it removed the whole string and just prints a newline.
So, what I need is to remove only the matched (.*?) part. So, that the output is
foo_bar__123.
There's another way, and it's quite simple:
my $str = "foo_bar_not_needed_string_part_123";
$str =~ s/(?<=foo_bar_)\D+//gi;
print $str;
The trick is to use lookbehind check anchor, and replace all non-digit symbols that follow this anchor (not a symbol). Basically, with this pattern you match only the symbols you need to be removed, hence no need for capturing groups.
As a sidenote, in the original regex (?=bar)bar construct is redundant. The first part (lookahead) will match only if some position is followed by 'bar' - but that's exactly what's checked with non-lookahead part of the pattern.
You can capture the parts you do not want to remove:
my $str = "foo_bar_not_needed_string_part_123";
$str =~ s/(foo_bar_).*?(_\d+)/$1$2/;
print $str;
You can try this:
my $str = "foo_bar_not_needed_string_part_123";
say $str if $str =~ s/(foo_(?=bar)bar_).*?(_\d+)/$1$2/;
Outputs:
foo_bar__123
PS: I am new to perl/regex so I am interested if there exist a way to directly replace the matched part. What I have done is captured everything which is required and than replaced the whole string with it.
What's about to divide string to 3 parts, and delete only middle?
$str =~ s/(foo_(?=bar)bar_)(.*?)(_\d+)/$1$3/;
Try this:
(?<=foo_bar_).*(?=_\d)
In this variant, it includes in result ALL (.*) between foo_bar_ and _"any digit".
In your regex, it includes in result:
foo_
Then it looks for "bar" after "foo_":
(?=bar)
But it DOES NOT included at this step. It is included on the next step:
bar_
And then rest of line is included by (.*?)_\d+.
So, in general: it includes in result all this that you typed, EXCEPT (?=bar), which is just looking for "bar" after expression.
go with
echo "foo_bar_not_needed_string_part_123" | perl -pe 's/(?<=foo_bar_)[^\d]+//'
You can use look-behind/look-ahead in this case
$str =~ s/(?<=foo_bar_).*?(?=_\d+)//;
and the look-behind can be replace with \K (keep) to make it a little tidier
$str =~ s/foo_bar_\K.*?(?=_\d+)//;

What does s-/-- and s-/\Z-- in perl mean?

I am a beginner in perl and I have a query regarding pattern matching.
I came across a line in perl where it was written
$variable =~ s-/\Z--;
And as the code goes ahead some another variable was assigned
$variable1 =~ s-/--;
Can you please tell me what does these 2 lines do?
I want to know what does s-/\Z-- and s-/-- mean.
$variable =~ s-/\Z--;
- is used as a delimiter here. However, best practice suggests that you either use / or {} as delimiters.
It could be re-written as:
$variable =~ s{/\Z}{}; # remove a / at the end of a string
Consider:
$variable1 =~ s-/--;
Again, it could be re-written as:
$variable1 =~ s{/}{}; # remove the first /
The s/// operator in Perl is a substitution operation, which performs a search-and-replace on a string using a special kind of pattern called a regular expression. You can read more about regular expressions and Perl's pattern matching in the man pages that come with Perl:
man perlretut
man perlre
If you don't have these on your system, try searching Google for the same.
Applying a substitution to a variable is done with the =~ operator. So the following replaces all instances of 'foo' in the variable $var with 'bar'.
$var =~ s/foo/bar/;
All the Perl operators are documented on the 'perlop' man page.
Even though the most common separator character is a slash (hence s///), you can also use any other punctuation character as a separator. So in this case, the author has decided to use the dash (-) as the separator.
Here's the same line of code above using dash as a separator:
$var =~ s-foo-bar-;
In your case, the dash doesn't seem to add any clarity to the code, so it might be best to update it to use the conventional slashes instead.
The s/// search and replace function in perl can be used with different delimeters, which is what is done in this case. They have replaced / with the minus sign -, or dash.
The s-/-- removes the first / from the string.
The s-/\Z-- matches and removes a slash at the end of the line. I think this is better written: s{/$}{}.
$variable1 =~ s-/--;could be written as
$variable =~ s{/}{}xms;
or this
$variable =~ s/ \/ //xms;
It means delete the first / in the string.
Regarding s-/\Z--, it is usually written like this
$variable =~ s{/ \Z}{}xms;
or this
$variable =~ s/ \/ \Z //xms;
It means delete a / if it is at the end of the string (\Z).

In Perl, how can I correctly extract URLs that are enclosed in parentheses?

I've got two question about Regexp::Common qw/URI/ and Regex in Perl.
I use Regexp::Common qw/URI/ to parse URI in the strings and delete them. But I've got an error when a URI is between parentheses.
For example: (http://www.example.com)
The error is caused by ')', and when it try to parse the URI, the app crash. So I've thought two fixes:
Do a simple (or I thought so) that writes a whitespace between parentheses and ) characters
The Regexp::Common qw/URI/ has a function that implement a fix.
In my code I've tried to implement the Regex but the app freezes. The code that I've tried is this:
use strict;
use Regexp::Common qw/URI/;
my $str = "Hello!!, I love (http://www.example.com)";
while ($str =~ m/\)/){
$str =~ s/\)/ \)/;
}
my ($uri) = $str =~ /$RE{URI}{-keep}/;
print "$uri\n";
print $str;
The output that I want is: (http://www.example.com )
I'm not sure, but I think that the problem is in $str =~ s/\)/ \)/;
BTW, I've got a question about Regexp::Common qw/URI/. I've got two string type:
ablalbalblalblalbal http://www.example.com
asfasdfasdf http://www.example.com aasdfasdfasdf
I want to remove the URI if it is the last component (and save it). And, if not, save it without removing it from the text.
You don't have to first test for a match to be able to use the s/// operator correctly: If the string does not match the search pattern, it will not do anything.
#!/usr/bin/perl
use strict; use warnings;
my $str = "Hello!!, I love (GOOGLE)";
$str =~ s/\)/ )/g;
print "$str\n";
The general problem of detecting URLs correctly in text is error-prone. See for example Jeff's thoughts on this.
my $str = "Hello!!, I love (GOOGLE)";
while ($str =~ m/)/){
$str =~ s/)/ )/;
}
Your program goes into an infinite loop at this point. To see why, try printing the value of $str each time round the loop.
my $str = "Hello!!, I love (GOOGLE)";
while ($str =~ m/)/){
$str =~ s/)/ )/;
print $str, "\n";
}
The first time it prints "Hello!!, I love (GOOGLE )". The while loop condition is then evaluated again. Your string still matches your regular expression (it still contains a closing parenthesis) so the replacement is run again and this time it prints out "Hello!!, I love (GOOGLE )" with two spaces.
And so it goes on. Each time round the loop another space is added, but each time you still have a closing parenthesis, so another substitution is run.
The simplest solution I can see is to only match the closing parenthesis if it is preceded by a non-whitespace character (using \S).
my $str = "Hello!!, I love (GOOGLE)";
while ($str =~ m/\S)/){
$str =~ s/)/ )/;
print $str, "\n";
}
In this case the loop is only executed once.
Why not just include the parentheses in the search? If the URLs will always be bracketed, then something like this:
#!/usr/bin/perl
use warnings;
use strict;
use Regexp::Common qw/URI/;
my $str = "Hello!!, I love (http://www.google.com)";
my ($uri) = $str =~ / \( ( $RE{URI} ) \) /x;
print "$uri\n";
The regex from Regex::Common can be used as part of a longer regex, it doesn't have to be used on its own. Also I've used the 'x' modifier on the regex to allow whitespace so you can see more clearly what is going on - the brackets with the backslashes are treated as characters to match, those without define what is to matched (presumably like the {-keep} - I've not used that before).
You could also make the brackets optional, with something like:
/ (?: \( ( $RE{URI} ) \) | ( $RE{URI} ) ) /
although that would result in two match variables, one undefined - so something like following would be needed:
my $uri = $1 || $2 || die "Didn't match a URL!";
There's probably a better way to do this, and also if you're not bothered about matching parentheses then you could simply make the brackets optional (via a '?') in the first regex...
To answer your second question about only matching URLs at the end of the line - have a look at Regex 'anchors' which can force a match against the beginning or end of a line: ^ and $ (or \A and \Z if you prefer). e.g. matching a URL at the end of a line only:
/$RE{URI}\Z/