I want to add a line at the beginning of every file in a directory.
perl -i.bkp -p -e 'print "#include top_level.reset\n" if $. == 1' *.reset
But this command is updating only the first file in the directory. I think this is because $. is not resetting for next file.
How to modify all the files.
You are correct, $. is not reset between files when processing #ARGV. See perlvar. You can work around it by explicitly closing ARGV on EOF - see eof. But I would not bother, instead I would use the shell to iterate the files:
for f in *.reset; do perl -i.bkp -p -e 'print "#include top_level.reset\n" if $. == 1' $f; done
ls -1 *.reset | xargs -n 1 perl -i.bkp -p -e 'print "#include top_level.reset\n" if $. == 1'
Related
I have a text file (input.txt) like this:
NP_414685.4: 15-26, 131-138, 441-465
NP_418580.2: 493-500
NP_418780.2: 36-48, 44-66
NP_418345.2:
NP_418473.3: 1-19, 567-1093
NP_418398.2:
I want a perl one-liner that keeps only those lines in file where ":" is followed by number range (that means, here, the lines containing "NP_418345.2:" and "NP_418398.2:" get deleted). For this I have tried:
perl -ni -e "print unless /: \d/" -pi.bak input.txt del input.txt.bak
But it shows exactly same output as the input file.
What will be the exact pattern that I can match here?
Thanks
First, print unless means print if not -- opposite to what you want.
More to the point, it doesn't make sense using both -n and -p, and when you do -p overrides the other. While both of them open the input file(s) and set up the loop over lines, -p also prints $_ for every iteration. So with it you are reprinting every line. See perlrun.
Finally, you seem to be deleting the .bak file ... ? Then don't make it. Use just -i
Altogether
perl -i -ne 'print if /:\s*\d+\s*-\s*\d+/' input.txt
If you do want to keep the backup file use -i.bak instead of -i
You can see the code equivalent to a one-liner with particular options with B::Deparse (via O module)
Try: perl -MO=Deparse -ne 1 and perl -MO=Deparse -pe 1
This way:
perl -i.bak -ne 'print if /:\s+\d+-\d/' input.txt
This:
perl -ne 'print if /:\s*(\d+\s*-\s*\d+\s*,?\s*)+\s*$/' input.txt
Prints:
NP_414685.4: 15-26, 131-138, 441-465
NP_418580.2: 493-500
NP_418780.2: 36-48, 44-66
NP_418473.3: 1-19, 567-1093
I'm not sure if you want to match lines that are possibly like this:
NP_418580.2: 493-500, asdf
or this:
NP_418580.2: asdf
This answer will not print these lines, if given to it.
What is the difference between the perl -n and perl -p options?
What is a simple example to demonstrate the difference?
How do you decide which one to use?
How do you decide which one to use?
You use -p if you want to automatically print the contents of $_ at the end of each iteration of the implied while loop. You use -n if you don't want to print $_ automatically.
An example of -p. Adding line numbers to a file:
$ perl -pe '$_ = "$.: $_"' your_file.txt
An example of -n. A basic grep replacement.
$ perl -ne 'print if /some search text/' your_file.txt
-p is short for -np, and it causes $_ to be printed for each pass of the loop created by -n.
perl -ne'...'
executes the following program:
LINE: while (<>) {
...
}
while
perl -pe'...'
executes the following program:
LINE: while (<>) {
...
}
continue {
die "-p destination: $!\n" unless print $_;
}
See perlrun for documentation about perl's command-line options.
What is the difference between the perl -n and perl -p options?
-p causes each line to be printed; equivalent to:
while (<>) { ... } continue { print }
-n does not automatically print each line; equivalent to:
while(<>) {...}
What is a simple example to demonstrate the difference?
e.g., replace foo with FOO:
$ echo 'foo bar' | perl -pe 's/foo/FOO/'
FOO bar
$ echo 'foo bar' | perl -ne 's/foo/FOO/'
$
How do you decide which one to use?
One example where -n is useful is when you don't want every line printed, and there is a conditional print in the code, e.g., only show lines containing foo:
$ echo -e 'foo\nbar\nanother foo' | perl -ne 'print if /foo/;'
foo
another foo
$
The command-line options are documented in perlrun documentation
perl -n is equivalent to while(<>){...}
perl -p is equivalent to while(<>){...;print;}
I am using Perl to search and replace multiple regular expressions:
When I execute the following command, I get an error:
prompt> find "*.cpp" | xargs perl -i -pe 's/##(\W)/\1/g' -pe 's/(\W)##/\1/g'
syntax error at -e line 2, near "s/(\W)##/\1/g"
Execution of -e aborted due to compilation errors.
xargs: perl: exited with status 255; aborting
Having multiple -e is valid in Perl, then why is this not working? Is there a solution to this?
Several -e's are allowed.
You are missing the ';'
find "*.cpp" | xargs perl -i -pe 's/##(\W)/\1/g;' -pe 's/(\W)##/\1/g;'
Perl statements has to end with ;.
Final statement in a block doesn't need a terminating semicolon.
So a single -e without ; will work, but you will have to add ; when you have multiple -e statements.
Having multiple -e values are valid, but is it useful? The values from the multiple -e are merely combined into one program, and it's up to you to ensure that together they make a syntactically correct program. The B::Deparse program can show you what perl thinks the program is:
$ perl -MO=Deparse -e 'print' -e 'q(Hello' -e ')'
print "Hello\n";
-e syntax OK
A curious thing to note is that a newline snuck in there. Think about how it got there to see what else perl is doing to combine multiple -e values.
In your program, you are substituting on the current line, then taking the modified line and substituting again. That's better written as:
prompt> find "*.cpp" | xargs perl -i -pe 's/##(\W)/\1/g; s/(\W)##/\1/g'
Now, if you are building up this command line by adding more and more -e through some automated process and you don't know ahead of time what you get, maybe those -e make sense. However, you might consider that you can do the same thing to build up the string you give to -e. I don't know what might be better because you didn't explain why you are doing it that way.
But, I suspect that in some cases, people are actually thinking about having only one substitution work. They want to try one and if its pattern doesn't work, try a different one until one succeeds. In that case you don't want to separate the substitutions by semicolons. Use the short-circuiting || instead. The s/// returns the number of substitutions it made and || will stop (short circuit) when it finds a true value:
prompt> find "*.cpp" | xargs perl -i -pe 's/##(\W)/\1/g || s/(\W)##/\1/g'
And note, you only need one -p. It only does its job once. Here's the program with multiple -p deparsed:
$ perl -MO=Deparse -i -pe 's/##(\W)/\1/g;' -pe 's/(\W)##/\1/g;'
BEGIN { $^I = ""; }
LINE: while (defined($_ = readline ARGV)) {
s/##(\W)/$1/g;
s/(\W)##/$1/g;
}
continue {
die "-p destination: $!\n" unless print $_;
}
-e syntax OK
It's the same thing as having only one -p:
$ perl -MO=Deparse -pi -e 's/##(\W)/\1/g;' -e 's/(\W)##/\1/g;'
BEGIN { $^I = ""; }
LINE: while (defined($_ = readline ARGV)) {
s/##(\W)/$1/g;
s/(\W)##/$1/g;
}
continue {
die "-p destination: $!\n" unless print $_;
}
-e syntax OK
Thanks so much! You helped me reduce my ascii / decimal / 8-bit binary table printer enough to fit in a tweet:
for i in {32..126};do printf "'\x$(printf %x $i)'(%3i) = " $i; printf '%03o\n' $i | perl \
-pe 's#0#000#g;' -pe 's#1#001#g;' -pe 's#2#010#g;' -pe 's#3#011#g;' \
-pe 's#4#100#g;' -pe 's#5#101#g;' -pe 's#6#110#g;' -pe 's#7#111#g' ; done | \
perl -pe 's#= 0#= #'
I want to delete all the rows/lines in a file that has a specific character, '?' in my case. I hope there is a single line command in Bash or AWK or Perl. Thanks
You can use sed to modify the file "in-place":
sed -i "/?/d" file
Alternatively, use grep:
grep -v "?" file > newfile.txt
Even better, just a single line using sed
sed '/?/d' input
use -i to edit file in place.
perl -i -ne'/\?/ or print' file
or
perl -i -pe's/^.*?\?.*//s' file
Here are already grep, sed and perl solutions - only for fun, pure bash one:
pattern='?'
while read line
do
[[ "$line" =~ "$pattern" ]] || echo "$line"
done
translated
for every line on the STDIN
match it for the pattern =~
and if the match is not successful || - print out the line
awk '!($0~/?/){print $0}' file_name
In a file, say xyz.txt i want to replace the pattern of any number followed by a dot example:1.,2.,10.,11. etc.. with a whitespace.
How to compose a perl command on the command line to act on the file to do the above, what should be the regex to be used ?
Please Help
Thank You.
This HAS to be a Perl oneliner?
perl -i -pe 's/\d+\./ /g' <fileName>
The Perl command line options: -i is used to specify what happens to the input file. If you don't give it a file extension, the original file is lost and is replaced by the Perl munged output. For example, if I had this:
perl -i.bak -pe 's/\d+\./ /g' <fileName>
The original file would be stored with a .bak suffix and <fileName> itself would contain your output.
The -p means to enclose your Perl program in a print loop that looks SOMEWHAT like this:
while ($_ = <>) {
<Your Perl one liner>
print "$_";
}
This is a somewhat simplified explanation what's going on. You can see the actual perl loop by doing a perldoc perlrun from the command line. The main idea is that it allows you to act on each line of a file just like sed or awk.
The -e simply contains your Perl command.
You can also do file redirection too:
perl -pe 's/\d+\./ /g' < xyz.txt > xyz.txt.out
Answer (not tested):
perl -ipe "s/\d+\./ /g" xyz.txt
Both
perl -ipe "s/\d+\./ /g" xyz.txt
and
perl -pie
cannot execute on my system.
I use the following order:
perl -i -pe