I have string
Message <Network=Data Center> All Verified
I need to extract all string except one in angular brackets
I tried
m/(?![^<]*\\>)/s
Not giving desired result.
Removing <..> regions
It's easier to remove the <..> parts from the string and then deal with the remaining string.
Try this oneliner:
cat file | perl -pne 's/<[^>]*?>//g;'
For your sample input, this is the output:
Message All Verified
Notice the non-greedy quantifier ? is used in the regex. Also, because this is a oneliner, the s/// search-and-replace construct is applied to $_ implicit variable (which is a line from standard input). So after search & replace has run in this oneliner, the $_ will be altered(there will be no <..> regions in it). Also the -p was used in order to print the variable $_ after running the block of code. You can read more about Perl commandline switches in perlrun.
This is one solution. Below there is another one:
Capturing regions outside of <..>
On the other hand, you can(if you want) match the parts outside of the <..> regions.
In order to do that let's build a regex. First, we want a < or > free region. The following regex matches just that
$p = ([^<>]*).
Next, we want to match everything before <, and for that we can write (?:$p<) and everything after >, and that's (?:>$p).
Now if we assemble all those parts together we get (?:>$p)|(?:$p<).
Notice that (?:) is a non-capturing group.
So now there are two capturing groups (the two $p you see above) but only one will match at a time, so some of the captures will be undef. We'll have to filter those out.
Finally, we can assemble all the captures, and we're done.
cat file | perl -ne '$p="([^<>]*)";#x=grep{defined} m{(?:>$p)|(?:$p<)}g; print join(" ",#x)."\n";'
Parse::Yapp parser
You might think that using Parser::Yapp for this particular problem is a bit too much(usually, if you have something complicated to parse, you would use a grammar and a parser generator), but .. why not.. :)
Ok, so we need a grammar, here's one right here grammar_file.yp:
#header
%%
#rules
expression:
| exterior '<' interior '>' exterior
| exterior
;
exterior:
| TOK { $_[0]->YYData->{DATA} .= $_[1]; }
| expression
;
interior: TOK;
%%
#footer
sub Error { my ($parser)=shift; }
sub Lexer {
use Data::Dumper;
my($parser)=shift;
$parser->YYData->{INPUT} or return('',undef);
#$parser->YYData->{INPUT}=~s/^\s+//;
for ($parser->YYData->{INPUT}) {
return ('TOK',$1) if(s/^([^<>]+)//);
return ( $1,$1) if(s/^([<>])//);
};
}
You will notice in the grammar above that the interior is completely ignored, and only the terminals from exterior are collected.
Here's a small program that will use the parser(MyParser.pm generated from grammar_file.yp) parse.pl:
#!/usr/bin/env perl
use strict;
use warnings;
use MyParser;
my $parser=MyParser->new;
$parser->YYData->{INPUT} = "Message <Network=Data Center> All Verified";
my $value=$parser->YYParse(
yylex => \&MyParser::Lexer,
yyerror => \&MyParser::Error,
#yydebug => 0x1F,
);
my $nberr=$parser->YYNberr();
my $data=$parser->YYData->{DATA};
print "Result=$data"
And now a Makefile and we're done:
generate_parser_module:
yapp -m MyParser grammar_file.yp;
run:
perl parse.pl
all: generate_parser_module
Note
Some more Parser generators can be found here
Regexp::Grammars
Parse::RecDescent
Marpa::XS or Marpa::R2
You can do it other way: just remove the string in the angular brackets:
s#<.*>##
Or if > is not allowed:
s#<[^>]*>##
You can use sed for that:
cat yourfile |sed 's/<.*>//g' > newfile
If you need perl:
perl -i -pe "s/<.*?>//g" yourfile
Here is a compact approach. The following regex will capture your strings into Group 1:
<[^>]+>|([^<>]*)
What we are interested in here is not the overall match, but just the Group 1 matches.
So we need to iterate over Group 1 matches. I don't code in Perl, but following a recipe from the perlretut tutorial, this should do it:
while ($x =~ /<[^>]+>|([^<>]*)/g) {
print "$1","\n";
}
Please give it a try and let me know if it works for you.
Related
I have a command that outputs text in the following format:
misc1=poiuyt
var1=qwerty
var2=asdfgh
var3=zxcvbn
misc2=lkjhgf
etc. I need to get the values for var1, var2, and var3 into variables in a perl script.
If I were writing a shell script, I'd do this:
OUTPUT=$(command | grep '^var-')
VAR1=$(echo "${OUTPUT}" | sed -ne 's/^var1=\(.*\)$/\1/p')
VAR2=$(echo "${OUTPUT}" | sed -ne 's/^var2=\(.*\)$/\1/p')
VAR3=$(echo "${OUTPUT}" | sed -ne 's/^var3=\(.*\)$/\1/p')
That populates OUTPUT with the basic content that I want (so I don't have to run the original command multiple times), and then I can pull out each value using sed VAR1 = 'qwerty', etc.
I've worked with perl in the past, but I'm pretty rusty. Here's the best I've been able to come up with:
my $output = `command | grep '^var'`;
(my $var1 = $output) =~ s/\bvar1=(.*)\b/$1/m;
print $var1
This correctly matches and references the value for var1, but it also returns the unmatched lines, so $var1 equals this:
qwerty
var2=asdfgh
var3=zxcvbn
With sed I'm able to tell it to print only the modified lines. Is there a way to do something similar with in perl? I can't find the equivalent of sed's p modifier in perl.
Conversely, is there a better way to extract those substrings from each line? I'm sure I could match match each line and split the contents or something like that, but was trying to stick with regex since that's how I'd typically solve this outside of perl.
Appreciate any guidance. I'm sure I'm missing something relatively simple.
One way
my #values = map { /\bvar(?:1|2|3)\s*=\s*(.*)/ ? $1 : () } qx(command);
The qx operator ("backticks") returns a list of all lines of output when used in list context, here imposed by map. (In a scalar context it returns all output in a string, possibly multiline.) Then map extracts wanted values: the ternary operator in it returns the capture, or an empty list when there is no match (so filtering out such lines). Please adjust the regex as suitable.
Or one can break this up, taking all output, then filtering needed lines, then parsing them. That allows for more nuanced, staged processing. And then there are libraries for managing external commands that make more involved work much nicer.
A comment on the Perl attempt shown in the question
Since the backticks is assigned to a scalar it is in scalar context and thus returns all output in a string, here multiline. Then the following regex, which replaces var1=(.*) with $1, leaves the next two lines since . does not match a newline so .* stops at the first newline character.
So you'd need to amend that regex to match all the rest so to replace it all with the capture $1. But then for other variables the pattern would have to be different. Or, could replace the input string with all three var-values, but then you'd have a string with those three values in it.
So altogether: using the substitution here (s///) isn't suitable -- just use matching, m//.
Since in list context the match operator also returns all matches another way is
my #values = qx(command) =~ /\bvar(?:1|2|3)\s*=\s*(.*)/g;
Now being bound to a regex, qx is in scalar context and so it returns a (here multiline) string, which is then matched by regex. With /g modifier the pattern keeps being matched through that string, capturing all wanted values (and nothing else). The fact that . doesn't match a newline so .* stops at the first newline character is now useful.
Again, please adjust the regex as suitable to yoru real problem.
Another need came up, to capture both the actual names of variables and their values. Then add capturing parens around names, and assign to a hash
my %val = map { /\b(var(?:1|2|3))\s*=\s*(.*)/ ? ($1, $2) : () } qx(command);
or
my %val = qx(command) =~ /\b(var(?:1|2|3))\s*=\s*(.*)/g;
Now the map for each line of output from command returns a pair of var-name + value, and a list of such pairs can be assigned to a hash. The same goes with subsequent matches (under /g) in the second case..
In scalar context, s/// and s///g return whether it found a match or not. So you can use
print $s if $s =~ s///;
I'm trying to return different replacement results with a perl regex one-liner if it matches a group. So far I've got this:
echo abcd | perl -pe "s/(ab)(cd)?/defined($2)?\1\2:''/e"
But I get
Backslash found where operator expected at -e line 1, near "1\"
(Missing operator before \?)
syntax error at -e line 1, near "1\"
Execution of -e aborted due to compilation errors.
If the input is abcd I want to get abcd out, if it's ab I want to get an empty string. Where am I going wrong here?
You used regex atoms \1 and \2 (match what the first or second capture captured) outside of a regex pattern. You meant to use $1 and $2 (as you did in another spot).
Further more, dollar signs inside double-quoted strings have meaning to your shell. It's best to use single quotes around your program[1].
echo abcd | perl -pe's/(ab)(cd)?/defined($2)?$1.$2:""/e'
Simpler:
echo abcd | perl -pe's/(ab(cd)?)/defined($2)?$1:""/e'
Simpler:
echo abcd | perl -pe's/ab(?!cd)//'
Either avoid single-quotes in your program[2], or use '\'' to "escape" them.
You can usually use q{} instead of single-quotes. You can also switch to using double-quotes. Inside of double-quotes, you can use \x27 for an apostrophe.
Why torture yourself, just use a branch reset.
Find (?|(abcd)|ab())
Replace $1
And a couple of even better ways
Find abcd(*SKIP)(*FAIL)|ab
Replace ""
Find (?:abcd)*\Kab
Replace ""
These use regex wisely.
There is really no need nowadays to have to use the eval form
of the regex substitution construct s///e in conjunction with defined().
This is especially true when using the perl command line.
Good luck...
I have a very large file, containing the following blocks of lines throughout:
start :234
modify 123 directory1/directory2/file.txt
delete directory3/file2.txt
modify 899 directory4/file3.txt
Each block starts with the pattern "start : #" and ends with a blank line. Within the block, every line starts with "modify # " or "delete ".
I need to modify the path in each line, specifically appending a directory to the front. I would just use a general regex to cover the entire file for "modify #" or "delete ", but due to the enormous amount of other data in that file, there will likely be other matches to this somewhat vague pattern. So I need to use multi-line matching to find the entire block, and then perform edits within that block. This will likely result in >10,000 modifications in a single pass, so I'm also trying to keep the execution down to less than 30 minutes.
My current attempt is a sed one-liner:
sed '/^start :[0-9]\+$/ { :a /^[modify|delete] .*$/ { N; ba }; s/modify [0-9]\+ /&Appended_DIR\//g; s/delete /&Appended_DIR\//g }' file_to_edit
Which is intended to find the "start" line, loop while the lines either start with a "modify" or a "delete," and then apply the sed replacements.
However, when I execute this command, no changes are made, and the output is the same as the original file.
Is there an issue with the command I have formed? Would this be easier/more efficient to do in perl? Any help would be greatly appreciated, and I will clarify where I can.
I think you would be better off with perl
Specifically because you can work 'per record' by setting $/ - if you're records are delimited by blank lines, setting it to \n\n.
Something like this:
#!/usr/bin/env perl
use strict;
use warnings;
local $/ = "\n\n";
while (<>) {
#multi-lines of text one at a time here.
if (m/^start :\d+/) {
s/(modify \d+)/$1 Appended_DIR\//g;
s/(delete) /$1 Appended_DIR\//g;
}
print;
}
Each iteration of the loop will pick out a blank line delimited chunk, check if it starts with a pattern, and if it does, apply some transforms.
It'll take data from STDIN via a pipe, or myscript.pl somefile.
Output is to STDOUT and you can redirect that in the normal way.
Your limiting factor on processing files in this way are typically:
Data transfer from disk
pattern complexity
The more complex a pattern, and especially if it has variable matching going on, the more backtracking the regex engine has to do, which can get expensive. Your transforms are simple, so packaging them doesn't make very much difference, and your limiting factor will be likely disk IO.
(If you want to do an in place edit, you can with this approach)
If - as noted - you can't rely on a record separator, then what you can use instead is perls range operator (other answers already do this, I'm just expanding it out a bit:
#!/usr/bin/env perl
use strict;
use warnings;
while (<>) {
if ( /^start :/ .. /^$/)
s/(modify \d+)/$1 Appended_DIR\//g;
s/(delete) /$1 Appended_DIR\//g;
}
print;
}
We don't change $/ any more, and so it remains on it's default of 'each line'. What we add though is a range operator that tests "am I currently within these two regular expressions" that's toggled true when you hit a "start" and false when you hit a blank line (assuming that's where you would want to stop?).
It applies the pattern transformation if this condition is true, and it ... ignores and carries on printing if it is not.
sed's pattern ranges are your friend here:
sed -r '/^start :[0-9]+$/,/^$/ s/^(delete |modify [0-9]+ )/&prepended_dir\//' filename
The core of this trick is /^start :[0-9]+$/,/^$/, which is to be read as a condition under which the s command that follows it is executed. The condition is true if sed currently finds itself in a range of lines of which the first matches the opening pattern ^start:[0-9]+$ and the last matches the closing pattern ^$ (an empty line). -r is for extended regex syntax (-E for old BSD seds), which makes the regex more pleasant to write.
I would also suggest using perl. Although I would try to keep it in one-liner form:
perl -i -pe 'if ( /^start :/ .. /^$/){s/(modify [0-9]+ )/$1Append_DIR\//;s/(delete )/$1Append_DIR\//; }' file_to_edit
Or you can use redirection of stdout:
perl -pe 'if ( /^start :/ .. /^$/){s/(modify [0-9]+ )/$1Append_DIR\//;s/(delete )/$1Append_DIR\//; }' file_to_edit > new_file
with gnu sed (with BRE syntax):
sed '/^start :[0-9][0-9]*$/{:a;n;/./{s/^\(modify [0-9][0-9]* \|delete \)/\1NewDir\//;ba}}' file.txt
The approach here is not to store the whole block and to proceed to the replacements. Here, when the start of the block is found the next line is loaded in pattern space, if the line is not empty, replacements are performed and the next line is loaded, etc. until the end of the block.
Note: gnu sed has the alternation feature | available, it may not be the case for some other sed versions.
a way with awk:
awk '/^start :[0-9]+$/,/^$/{if ($1=="modify"){$3="newdirMod/"$3;} else if ($1=="delete"){$2="newdirDel/"$2};}{print}' file.txt
This is very simple in Perl, and probably much faster than the sed equivalent
This one-line program inserts Appended_DIR/ after any occurrence of modify 999 or delete at the start of a line. It uses the range operator to restrict those changes to blocks of text starting with start :999 and ending with a line containing no printable characters
perl -pe"s<^(?:modify\s+\d+|delete)\s+\K><Appended_DIR/> if /^start\s+:\d+$/ .. not /\S/" file_to_edit
Good grief. sed is for simple substitutions on individual lines, that is all. Once you start using constructs other than s, g, and p (with -n) you are using the wrong tool. Just use awk:
awk '
/^start :[0-9]+$/ { inBlock=1 }
inBlock { sub(/^(modify [0-9]+|delete) /,"&Appended_DIR/") }
/^$/ { inBlock=0 }
{ print }
' file
start :234
modify 123 Appended_DIR/directory1/directory2/file.txt
delete Appended_DIR/directory3/file2.txt
modify 899 Appended_DIR/directory4/file3.txt
There's various ways you can do the above in awk but I wrote it in the above style for clarity over brevity since I assume you aren't familiar with awk but should have no trouble following that since it reuses your own sed scripts regexps and replacement text.
I coded perl regex to extract the words after a certain anchor,
it seems like its not working. What am I doing wrong.
This is my actual output, I need to extract every number after groups keyword
$id cuser301 uid=2301(cuser301) gid=32(rpc) groups=32(rpc),1001(cgrp1),1002(cgrp2),1003(cgrp3),1004(cgrp4),1005(cgrp5),1006(cgrp6),1007(cgrp7),1008(cgrp8),1009(cgrp9),1010(cgrp10),1011(cgrp11),1012(cgrp12),1013(cgrp13),1014(cgrp14),1015(cgrp15),1016(cgrp16),1017(cgrp17),1018(cgrp18),1019(cgrp19),1020(cgrp20),1021(cgrp21),1022(cgrp22),1023(cgrp23),1024(cgrp24),1025(cgrp25),1026(cgrp26),1027(cgrp27),1028(cgrp28),1029(cgrp29),1030(cgrp30),1031(cgrp31),1032(cgrp32)
From the above, I run the id command and then would like to capture the numbers after groups Please help.
I am using the following.
my $check_groups = execute("\id $user"); #---> (execute is to run commands on the linux client, please ignore it)
my $new_groups = ('/^groups/',$check_groups); # ---> Now $new_groups should have all numbers after groups.
my $input = '$id cuser301 uid=2301(cuser301) gid=32(rpc) groups=32(rpc),1001(cgrp1),1002(cgrp2),1003(cgrp3),1004(cgrp4),1005(cgrp5),1006(cgrp6),1007(cgrp7),1008(cgrp8),1009(cgrp9),1010(cgrp10),1011(cgrp11),1012(cgrp12),1013(cgrp13),1014(cgrp14),1015(cgrp15),1016(cgrp16),1017(cgrp17),1018(cgrp18),1019(cgrp19),1020(cgrp20),1021(cgrp21),1022(cgrp22),1023(cgrp23),1024(cgrp24),1025(cgrp25),1026(cgrp26),1027(cgrp27),1028(cgrp28),1029(cgrp29),1030(cgrp30),1031(cgrp31),1032(cgrp32)';
print join ',', $input =~ /(?:.*groups=|\G.*?)\b([0-9]+)/g;
This is a common pattern; in more complicated cases where you want to ensure the \G branch only applies after the first non-zero-length match, you can use \G(?!\A) instead of just \G.
Try doing this :
$ echo <INPUT> | perl -ne 'print "$1," while /,(\d+)\(/g'
Check https://regex101.com/r/uZ9tO6/1
How do I remove duplicate characters and keep the unique one only.
For example, my input is:
EFUAHUU
UUUEUUUUH
UJUJHHACDEFUCU
Expected output is:
EFUAH
UEH
UJHACDEF
I came across perl -pe's/$1//g while/(.).*\/' which is wonderful but it is removing even the single occurrence of the character in output.
This can be done using positive lookahead :
perl -pe 's/(.)(?=.*?\1)//g' FILE_NAME
The regex used is: (.)(?=.*?\1)
. : to match any char.
first () : remember the matched
single char.
(?=...) : +ve lookahead
.*? : to match anything in between
\1 : the remembered match.
(.)(?=.*?\1) : match and remember
any char only if it appears again
later in the string.
s/// : Perl way of doing the
substitution.
g: to do the substitution
globally...that is don't stop after
first substitution.
s/(.)(?=.*?\1)//g : this will
delete a char from the input string
only if that char appears again later
in the string.
This will not maintain the order of the char in the input because for every unique char in the input string, we retain its last occurrence and not the first.
To keep the relative order intact we can do what KennyTM tells in one of the comments:
reverse the input line
do the substitution as before
reverse the result before printing
The Perl one line for this is:
perl -ne '$_=reverse;s/(.)(?=.*?\1)//g;print scalar reverse;' FILE_NAME
Since we are doing print manually after reversal, we don't use the -p flag but use the -n flag.
I'm not sure if this is the best one-liner to do this. I welcome others to edit this answer if they have a better alternative.
if Perl is not a must, you can also use awk. here's a fun benchmark on the Perl one liners posted against awk. awk is 10+ seconds faster for a file with 3million++ lines
$ wc -l <file2
3210220
$ time awk 'BEGIN{FS=""}{delete _;for(i=1;i<=NF;i++){if(!_[$i]++) printf $i};print""}' file2 >/dev/null
real 1m1.761s
user 0m58.565s
sys 0m1.568s
$ time perl -n -e '%seen=();' -e 'for (split //) {print unless $seen{$_}++;}' file2 > /dev/null
real 1m32.123s
user 1m23.623s
sys 0m3.450s
$ time perl -ne '$_=reverse;s/(.)(?=.*?\1)//g;print scalar reverse;' file2 >/dev/null
real 1m17.818s
user 1m10.611s
sys 0m2.557s
$ time perl -ne'my%s;print grep!$s{$_}++,split//' file2 >/dev/null
real 1m20.347s
user 1m13.069s
sys 0m2.896s
perl -ne'my%s;print grep!$s{$_}++,split//'
Here is a solution, that I think should work faster than the lookahead one, but is not regexp-based and uses hashtable.
perl -n -e '%seen=();' -e 'for (split //) {print unless $seen{$_}++;}'
It splits every line into characters and prints only the first appearance by counting appearances inside %seen hashtable
Tie::IxHash is a good module to store hash order (but may be slow, you will need to benchmark if speed is important). Example with tests:
use Test::More 0.88;
use Tie::IxHash;
sub dedupe {
my $str=shift;
my $hash=Tie::IxHash->new(map { $_ => 1} split //,$str);
return join('',$hash->Keys);
}
{
my $str='EFUAHUU';
is(dedupe($str),'EFUAH');
}
{
my $str='EFUAHHUU';
is(dedupe($str),'EFUAH');
}
{
my $str='UJUJHHACDEFUCU';
is(dedupe($str),'UJHACDEF');
}
done_testing();
Use uniq from List::MoreUtils:
perl -MList::MoreUtils=uniq -ne 'print uniq split ""'
If the set of characters that can be encountered is restricted, e.g. only letters, then the easiest solution will be with tr
perl -p -e 'tr/a-zA-Z/a-zA-Z/s'
It will replace all the letters by themselves, leaving other characters unaffected and /s modifier will squeeze repeated occurrences of the same character (after replacement), thus removing duplicates
Me bad - it removes only adjoining appearances. Disregard
This looks like a classic application of positive lookbehind, but unfortunately perl doesn't support that. In fact, doing this (matching the preceding text of a character in a string with a full regex whose length is indeterminable) can only be done with .NET regex classes, I think.
However, positive lookahead supports full regexes, so all you need to do is reverse the string, apply positive lookahead (like unicornaddict said):
perl -pe 's/(.)(?=.*?\1)//g'
And reverse it back, because without the reverse that'll only keep the duplicate character at the last place in a line.
MASSIVE EDIT
I've been spending the last half an hour on this, and this looks like this works, without the reversing.
perl -pe 's/\G$1//g while (/(.).*(?=\1)/g)' FILE_NAME
I don't know whether to be proud or horrified. I'm basically doing the positive looakahead, then substituting on the string with \G specified - which makes the regex engine start its matching from the last place matched (internally represented by the pos() variable).
With test input like this:
aabbbcbbccbabb
EFAUUUUH
ABCBBBBD
DEEEFEGGH
AABBCC
The output is like this:
abc
EFAUH
ABCD
DEFGH
ABC
I think it's working...
Explanation - Okay, in case my explanation last time wasn't clear enough - the lookahead will go and stop at the last match of a duplicate variable [in the code you can do a print pos(); inside the loop to check] and the s/\G//g will remove it [you don't need the /g really]. So within the loop, the substitution will continue removing until all such duplicates are zapped. Of course, this might be a little too processor intensive for your tastes... but so are most of the regex-based solutions you'll see. The reversing/lookahead method will probably be more efficient than this, though.
From the shell, this works:
sed -e 's/$/<EOL>/ ; s/./&\n/g' test.txt | uniq | sed -e :a -e '$!N; s/\n//; ta ; s/<EOL>/\n/g'
In words: mark every linebreak with a <EOL> string, then put every character on a line of its own, then use uniq to remove duplicate lines, then strip out all the linebreaks, then put back linebreaks instead of the <EOL> markers.
I found the -e :a -e '$!N; s/\n//; ta part in a forum post and I don't understand the seperate -e :a part, or the $!N part, so if anyone can explain those, I'd be grateful.
Hmm, that one does only consecutive duplicates; to eliminate all duplicates you could do this:
cat test.txt | while read line ; do echo $line | sed -e 's/./&\n/g' | sort | uniq | sed -e :a -e '$!N; s/\n//; ta' ; done
That puts the characters in each line in alphabetical order though.
use strict;
use warnings;
my ($uniq, $seq, #result);
$uniq ='';
sub uniq {
$seq = shift;
for (split'',$seq) {
$uniq .=$_ unless $uniq =~ /$_/;
}
push #result,$uniq;
$uniq='';
}
while(<DATA>){
uniq($_);
}
print #result;
__DATA__
EFUAHUU
UUUEUUUUH
UJUJHHACDEFUCU
The output:
EFUAH
UEH
UJHACDEF
for a file containing the data you list named foo.txt
python -c "print set(open('foo.txt').read())"