I'm new to Perl and I have been learning about the Perl basics for past two days.
I'm converting a Perl script to Java program gradually.
In the Perl script, I came across this code.
if( $arr[$i]=~/^0$/ ){
...
...
}
I know that $arr[$i] means getting the ith element from the array arr.
But what does =~/^0$/ mean?
To what are they comparing the array's element?
I searched for this, but I couldn't find it.
Someone please explain me.
FYI, the arr contains floating values.
if ($arr[$i]) =~ /^0$/) is roughly equivalent to if ($arr[$i] eq "0"), but not exactly the same, as it will match both the strings "0" and "0\n". If $arr[$1] was read from a file or stdin and it has not been chomped, this can be a very significant distinction.
if ($arr[$i] == 0), on the other hand, will match any string beginning with a non-numeric character or a string of zeroes/whitespace which is not followed by a numeric character, although it will generate a warning if the string contains non-whitespace, non-digit characters or contains only whitespace (and warnings are enabled, of course).
=~ is a binding operator.
"Binary "=~" binds a scalar expression to a pattern match"
/^0$/ on the right hand side is the regex
^ Match the beginning of the line
$ Match the end of the line (or before newline at the end)
And the zero has no special meaning.
^ and $ are regex anchors which says $arr[$i] should begin with 0 and there is end of string immediately after it.
It can be written as
if ($arr[$i] eq "0" or $arr[$i] eq "0\n")
Related
(my $batch_name = $batch_dir) =~ s#.*/##;
I had come across this statement while going through a script and tried to understand it. Even googling the RHS did not return anything useful. Can someone please help me understand what this statement means???
which out of the two scalars are affected??
it deletes the longest prefix ending with a / from a copy of the $batch_dir variable, eg. producing a leafname from a file system path or extracting the script, query and fragment part of a properly escaped url.
the idiom actually comprises 2 operations:
my $batch_name = $batch_dir;
batch_name =~ s#.*/##;
without the parentheses the substitution would be applied to $batch_dir and $batch_name would be set to the value returned from the substitution operator, the success status (at least 1 substitution has occurred => 1, undef else).
Parens () has higher priority than =~, so the instruction inside parens is executed before.
First the assignement my $batch_name = $batch_dir; is done, then the substitution $batch_name =~ s#.*/##;
Only the $batch_name variable is affected by the substitution.
Here is a piece of code
while($l=~/(\\\s*)$/) {
statements;
}
$l contains a line of text taken form file, in effect this code is for go through lines in file.
Questions:
I don't clearly understand what the condition in while is doing. I think it is trying to match group of \ followed by some number of white spaces at the end of line and loop should stop whenever a line ends with \ and may be some white spaces. I am not sure of it.
I came across statement $a ~= s/^(.*$)/$1/ . What I understand that ^ will force matching at the beginning of string, but in (.*$) would mean match all the characters at the end of string . Dose it mean that the statement is trying to find if any group of character at the end is same as group of character in the beginning of text ?
It is interesting to note that this statement:
while ( $l =~ /(\\\s*)$/ ) {
Is an infinite loop unless $l is altered inside the loop so that the regex no longer matches. As has already been mentioned by others, this is what it matches:
( ... ) a capture group, captures string to $1 (that's the number one, not lower case L)
\\ matches a literal backslash
\s* matches 0 or more whitespace characters.
$ matches end of line with optional newline.
Since you do not have the /g modifier, this regex will not iterate through matches, it will simply check if there is a match, resetting the regex each iteration, thereby causing an endless loop.
The statement
$a ~= s/^(.*$)/$1/
Looks rather pointless. It captures a string of characters up until end of string, then replaces it with itself. The captured text is stored in $1 and is simply replaced. The only marginally useful thing about this regex is that:
It matches up until newline \n, and nothing further, which may be of some use to a parser. A period . matches any character except newline, unless the /s modifier is present on the regex.
It captures the line in $1 for future use. However, a simple /^(.*$)/ would do the same.
1. the while
Usually while (regex) is used with the /g modifier, otherwise, if it matches, you get an infinite loop (unless you exit the loop, like using last).
statements would be executed continuously in an infinite loop.
In your case, adding the g
while($l=~/(\\\s*)$/g)
will have the while make only one loop, due to the $ - making a match unique (whatever matches up to the end of string is unique, as $ marks the end, and there is nothing after...).
2. $a ~= s/^(.*$)/$1/
This is a substitution. If the string ^.*$ matches (and it will, since ^.*$ matches (almost, see comment) anything) it is replaced with... $1 or what's inside the (), ie itself, since the match occurs from 1st char to the end of string
^ means beginning of string
(.*) means all chars
$ end of string
so that will replace $a with itself - probably not what you want.
it matches a literal backslash followed by 0 or more spaces followed by the end of the line.
it executes statements for all the lines in that text file that contain a \, followed by zero or more spaces ( \s* ), at the end of the line ($).
It matches lines that end with a backslash character, ignoring any trailing whitespace characters.
Ending a line with a backslash is used in some languages and data files to indicate that the line is being continued on the next line. So I suspect this is part of a parser that merges these continuation lines.
If you enter a regular expression at RegExr and hover your mouse over the pieces, it displays the meaning of each piece in a tooltip.
(\\\s*)$ this regex means --- a \ followed by zero or more number of white space characters which is followed by end of the line. Since you have your regex in (...), you can extract what you matched using $1, if you need.
http://rubular.com/r/dtHtEPh5DX
EDIT -- based on your update
$a ~= s/^(.$)/$1/ --- this is search and replace. So your regex matches a line which contains exactly one character (since you use . http://www.regular-expressions.info/dot.html), except a new-line character. Since you use (...), the character which matched the regex is extracted and stored in variable a
EDIT -- you changed your regex so here is the updated answer
$a ~= s/^(.*$)/$1/ -- same as above except now it matches zero or more characters (except new-line)
I've been googling for some time now and strangely didn't find anything that answers my question.
I want to pass -n as an option to the program, where n is an integer.
This is what I have:
if($ARGV[0] eq "-A_NUMBER")
Is there some type of wildcard I can use for this? So the condition is true for any minus sign followed by any integer (or character)?
I'd reccomend having a read about Getopt::Std it is more long winded than just inspecting #ARGV, but more robust
Use
if ($ARGV[0] =~ /^-\d/)
This regular expression matches a minus sign then a number.
"^" anchors the match to the start of the line and "\d" is an escape character that represents a number
I've to match a regular-expression, stored in a variable:
#!/bin/env perl
use warnings;
use strict;
my $expr = qr/\s*(\w+(\[\d+\])?)\s+(\w+(\[\d+\])?)/sx;
$str = "abcd[3] xyzg[4:0]";
if ($str =~ m/$expr/) {
print "\n%%%%%%%%% $`-----$&-----$'\n";
}
else {
print "\n********* NOT MATCHED\n";
}
But I'm getting the outout in $& as
%%%%%%%%% -----abcd[3] xyzg-----[4:0]
But expecting, it shouldn't go inside the if clause.
What is intended is:
if $str = "abcd xyzg" => %%%%%%%%% -----abcd xyzg----- (CORRECT)
if $str = "abcd[2] xyzg" => %%%%%%%%% -----abcd[2] xyzg----- (CORRECT)
if $str = "abcd[2] xyzg[3] => %%%%%%%%% -----abcd[2] xyzg[3]----- (CORRECT)
if $str = "abcd[2:0] xyzg[3] => ********* NOT MATCHED (CORRECT)
if $str = "abcd[2:0] xyzg[3:0] => ********* NOT MATCHED (CORRECT)
if $str = "abcd[2] xyzg[3:0]" => ********* NOT MATCHED (CORRECT/INTENDED)
but output is %%%%%%%%% -----abcd[2] xyzg-----[3:0] (WRONG)
OR better to say this is not intended.
In this case, it should/my_expectation go to the else block.
Even I don't know, why $& take a portion of the string (abcd[2] xyzg), and $' having [3:0]?
HOW?
It should match the full, not a part like the above. If it didn't, it shouldn't go to the if clause.
Can anyone please help me to change my $expr pattern, so that I can have what is intended?
By default, Perl regexes only look for a matching substring of the given string. In order to force comparison against the entire string, you need to indicate that the regex begins at the beginning of the string and ends at the end by using ^ and $:
my $expr = qr/^\s*(\w+(\[\d+\])?)\s+(\w+(\[\d+\])?)$/;
(Also, there's no reason to have the /x modifier, as your regex doesn't include any literal whitespace or # characters, and there's no reason for the /s modifier, as you're not using ..)
EDIT: If you don't want the regex to match against the entire string, but you want it to reject anything in which the matching portion is followed by something like "[0:0]", the simplest way would be to use lookahead:
my $expr = qr/^\s*(\w+(\[\d+\])?)\s+(\w+(\[\d+\]|(?=[^[\w])|$ ))/x;
This will match anything that takes the following form:
beginning of the string (which your example in the comments seems to imply you want)
zero or more whitespace characters
one or more word characters
optional: [, one or more digits, ]
one or more whitespace characters
one or more word characters
one of the following, in descending order of preference:
[, one or more digits, ]
an empty string followed by (but not including!) a character that is neither [ nor a word character (The exclusion of word characters is to keep the regex engine from succeeding on "a[0] bc[1:2]" by only matching "a[0] b".)
end of string (A space is needed after the $ to keep it from merging with the following ) to form the name of a special variable, and this entails the reintroduction of the /x option.)
Do you have any more unstated requirements that need to be satisfied?
The short answer is your regexp is wrong.
We can't fix it for you without you explaining what you need exactly, and the community is not going to write a regexp exactly for your purpose because that's just too localized a question that only helps you this one time.
You need to ask something more general about regexps that we can explain to you, that will help you fix your regexp, and help others fix theirs.
Here's my general answer when you're having trouble testing your regexp. Use a regexp tool, like the regex buddy one.
So I'm going to give a specific answer about what you're overlooking here:
Let's make this example smaller:
Your pattern is a(bc+d)?. It will match: abcd abccd etc. While it will not match bcd nor bzd in the case of abzd it will match as matching only a because the whole group of bc+d is optional. Similarly it will match abcbcd as a dropping the whole optional group that couldn't be matched (at the second b).
Regexps will match as much of the string as they can and return a true match when they can match something and have satisfied the entire pattern. If you make something optional, they will leave it out when they have to including it only when it's present and matches.
Here's what you tried:
qr/\s*(\w+(\[\d+\])?)\s+(\w+(\[\d+\])?)/sx
First, s and x aren't needed modifiers here.
Second, this regex can match:
Any or no whitespace followed by
a word of at least one alpha character followed by
optionally a grouped square bracketed number with at least one digit (eg [0] or [9999]) followed by
at least one white space followed by
a word of at least one alpha character followed by
optionally a square bracketed number with at least one digit.
Clearly when you ask it to match abcd[0] xyzg[0:4] the colon ends the \d+ pattern but doesn't satisfy the \] so it backtracks the whole group, and then happily finds the group was optional. So by not matching the last optional group, your pattern has matched successfully.
I am parsing a log file trying to pull out the lines where the phrase "failures=" is a unique non-zero digit.
The first part of my perl one liner will pull out all the lines where "failures" are greater than zero. But that part of the log file repeats until a new failure occurs, i.e., after the first failure the log entries will be "failures=1" until the second error then it will read, "failures=2".
What I'd like to do is pull only the first line where that value changes and I thought I had it with this:
cat -n Logstats.out | perl -nle 'print "Line No. $1: failures=$2; eventDelta=$3; tracking_id=$4" if /\s(\d+)\t.*failures=(\d+).*eventDelta=(.\d+).*tracking_id="(\d+)"/ && $2 > 0' | perl -ne 'print unless $a{/failures=\d+/}++'
However, that only pulls the first non-zero "failure" line and nothing else. What do I need to change for it to pull all the unique values for "failures"?
thanks in advance!
Update: The amount of text in each line up to the "tracking_id" is more text than I can post. Sorry!
2011-09-06 14:14:18 [INFO] [logStats]: name=somename id=d6e6f4f0-4c0d-93b6-7a71-8e3100000030
successes=1 failures=0 skipped=0 eventDelta=41 original=188 simulated=229
totalDelta=41 averageDelta=41 min=0 max=41 averageOriginalDuration=188 averageSimulatedDuriation=229(txid = b3036eca-6288-48ef-166f-5ae200000646
date = 2011-09-02 08:00:00 type = XML xml
=
perl -ne 'print unless $a{/failures=\d+/}++'
does not work because a hash subscript is evaluated in scalar context, and the m// operator does not return the match in scalar context. Instead, it returns 1. So (since every line matches), what you wrote is equivalent to:
perl -ne 'print unless $a{1}++'
and I think you can see the problem there.
There's a number of ways to work around that, but I'd use:
perl -ne 'print unless /(failures=\d+)/ and $a{$1}++'
However, I'd do the whole thing in one call to perl, including numbering the lines:
perl -nle '
print "Line No. $.: failures=$1; eventDelta=$2; tracking_id=$3"
if /failures=(\d+).*?eventDelta=(.\d+).*?tracking_id="(\d+)"/
&& $1 > 0
&& !$seen{$1}++
' Logstats.out
($. automatically counts input lines in Perl. The line breaks can be removed if desired, but it will actually work as is.)
you could use a hash to store te results and print it:
perl -nle '$f{$2} ||= "Line No. $1: failures=$2; eventDelta=$3; tracking_id=$4" if /\s(\d+)\t.*failures=(\d+).*eventDelta=(.\d+ ).*tracking_id="(\d+)"/ && $2;END{ print $f{$_} for keys %f }' Logstats.out
(not tested due to missing input data...)
HTH,
Paul
Since your input does not match your regex, I can't really help you. But I can tell you that this is doing a lot of backtracking--and that's bad if there is a lot of data after the part that you're interested in.
So here is some alternative ideas:
qr{ \s # a single space
failures=(\d+) # the entry for failures
\s+ # at least one space
skipped=\d+ # skipped
\s+
eventDelta=(.\d+)
.*? # any number of non-newline characters *UNTIL* ...
\btracking_id="(\d+)" # another specified sequence of data
}x;
The parser will scan "skipped=" and then a group of digits a lot faster than scanning the rest of the line and backtracking when it fails back to 'eventDelta=', it is better to put it in, if you know it will always be there.
Since you don't put tracking_id in your example, I can't tell how it occurs, so in this case we used a non-greedy any match which will always be looking for the next sequence. Again, if there is a lot of data in the line, then you do not want to scan to then end and backtrack until you find that you've already read 'tracking_id="nnn"'. However, lookaheads cost processing time, it is still better to spell out 'skipped=' and all possible values then a non-greedy "any match".
You'll also notice that after accepting any data, I specify that tracking_id should appear at a word boundary, which disambiguates it from the possible--though not likely 'backtracking_id='.