Why does the following not replace multiple empty lines with one?
$ cat some_random_text.txt
foo
bar
test
and this does not work:
$ cat some_random_text.txt | perl -pe "s/\n+/\n/g"
foo
bar
test
I am trying to replace the multiple new lines (i.e. empty lines) to a single empty new line but the regex I use for that does not work as you can see in the example snippet.
What am I messing up?
Expected outcome is:
foo
bar
test
The reason it doesn't work is that -p tells perl to process the input line by line, and there's never more than one \n in a single line.
Better idea:
perl -00 -lpe 1
-00: Enable paragraph mode (input records are terminated by any sequence of 2+ newlines).
-l: Enable autochomp mode (the input record separators are trimmed automatically, so since we're in paragraph mode, all trailing newlines are removed, and output records get "\n\n" added).
-p: Enable automatic input/output (the main code is executed for each input record; anything left in $_ is printed automatically).
-e 1: Use a dummy main program that does nothing.
Taken all together this does nothing except normalize paragraph terminators to exactly two newlines.
You are executing the following program:
LINE: while (<>) {
s/\n+/\n/g;
}
continue {
die "-p destination: $!\n" unless print $_;
}
Since you are reading one line at at time, and since a line is a sequence of characters that aren't line feeds terminated by a line feed, your pattern will never match more than one newline.
The simple fix is to tell Perl to treat the entire file as one line. Also, you don't want to replace every line feed, but just those found in sequence of two or more, and you want to replace the sequence with two line feeds.
perl -0777pe's/\n\n\K\n+//g; s^\n+//; s/\n\K\n\z//' some_random_text.txt
The second and third substitutions ensure there are no blank lines at the start and end of the file.
While reading the entire file into memory is easy, it's not necessary. The desired output can also be achieved by maintaining a flag that indicates whether the previous line was blank or not.
perl -ne'if (/\S/) { print "\n" if $f; print; $f=0 } else { $f=1 }' some_random_text.txt
This solution also removes blank lines from the start and end of the file.
Given:
$ echo "$txt"
foo
bar
test
You can use sed to reduce the runs of blank lines to a single \n:
$ echo "$txt" | sed '/^$/N;/^\n$/D'
foo
bar
test
Even easier, you can use cat -s:
$ echo "$txt" | cat -s # same output
In perl the idiomatic 1 liner is to use -00 for paragraph mode:
$ echo "$txt" | perl -00pe0 # same output
And in awk you have the flexibility of using paragraph mode by setting RS= and then set ORS= to what you want the replacement for runs of \n to be:
$ echo "$txt" | awk '1' RS= ORS="\n\n" # same output
ikegami correctly states that printf 'a\n\n' | ... will produce two trailing spaces with these solutions. That may or may not be an issue.
Related
I want to delete the header from all the files, and the header has the lines starting with //.
If I want to delete all the lines that starts with //, I can do following:
sed '/^\/\//d'
But, that is not something I need to do. I just need to delete the lines in the beginning of the file that starts with //.
Sample file:
// This is the header
// This should be deleted
print "Hi"
// This should not be deleted
print "Hello"
Expected output:
print "Hi"
// This should not be deleted
print "Hello"
Update:
If there is a new line in the beginning or in-between, it doesn't work. Is there any way to take care of that scenario?
Sample file:
< new empty line >
// This is the header
< new empty line >
// This should be deleted
print "Hi"
// This should not be deleted
print "Hello"
Expected output:
print "Hi"
// This should not be deleted
print "Hello"
Can someone suggest a way to do this? Thanks in advance!
Update: The accepted answer works well for white space in the beginning or in-between.
Could you please try following. This also takes care of new line scenario too, written and tested in https://ideone.com/IKN3QR
awk '
(NF == 0 || /^[[:blank:]]*\/\//) && !found{
next
}
NF{
found=1
}
1
' Input_file
Explanation: Simply checking conditions if a line either is empty OR starting from // AND variable found is NULL then simply skip those lines. Once any line without // found then setting variable found here so all next coming lines should be printed from line where it's get set to till end of Input_file printed.
With sed:
sed -n '1{:a; /^[[:space:]]*\/\/\|^$/ {n; ba}};p' file
print "Hi"
// This should not be deleted
print "Hello"
Slightly shorter version with GNU sed:
sed -nE '1{:a; /^\s*\/\/|^$/ {n; ba}};p' file
Explanation:
1 { # execute this block on the fist line only
:a; # this is a label
/^\s*\/\/|^$/ { n; # on lines matching `^\s*\/\/` or `^$`, do: read the next line
ba } # and go to label :a
}; # end block
p # print line unchanged:
# we only get here after the header or when it's not found
sed -n makes sed not print any lines without the p command.
Edit: updated the pattern to also skip empty lines.
I sounds like you just want to start printing from the first line that's neither blank nor just a comment:
$ awk 'NF && ($1 !~ "^//"){f=1} f' file
print "Hi"
// This should not be deleted
print "Hello"
The above simply sets a flag f when it finds such a line and prints every line from then on. It will work using any awk in any shell on every UNIX box.
Note that, unlike some of the potential solutions posted, it doesn't store more than 1 line at a time in memory and so will work no matter how large your input file is.
It was tested against this input:
$ cat file
// This is the header
// This should be deleted
print "Hi"
// This should not be deleted
print "Hello"
To run the above on many files at once and modify each file as you go is this with GNU awk:
awk -i inplace 'NF && ($1 !~ "^//"){f=1} f' *
and this with any awk:
ip_awk() { local f t=$(mktemp) && for f in "${#:2}"; do awk "$1" "$f" > "$t" && mv -- "$t" "$f"; done; }
ip_awk 'NF && ($1 !~ "^//"){f=1} f' *
In case perl is available then this may also work in slurp mode:
perl -0777 -pe 's~\A(?:\h*(?://.*)?\R+)+~~' file
\A will only match start of the file and (?:\h*(?://.*)?\R+)+ will match 1 or more lines that are blank or have // with optional leading spaces.
With GNU sed:
sed -i -Ez 's/^((\/\/[^\n]*|\s*)\n)+//' file
The ^((\/\/[^\n]*|\s*)\n)+ expression will match one or more lines starting with //, also matching blank lines, only at the start of the file.
Using ed (the file editor that the stream editor sed is based on),
printf '1,/^[^/]/ g|^\(//.*\)\{0,1\}$| d\nw\n' | ed tmp.txt
Some explanations are probably in order.
ed takes the name of the file to edit as an argument, and reads commands from standard input. Each command is terminated by a newline. (You could also read commands from a here document, rather than from printf via a pipe.)
1,/^[^/]/ addresses the first lines in the file, up to and including the first one that does not start with /. (All the lines you want to delete will be included in this set.)
g|^\(//.*\)\{0,1\}$|d deletes all the addressed lines that are either empty or do start with //.
w saves the changes.
Step 2 is a bit ugly; unfortunately, ed does not support regular expression operators you may take for granted, like ? or |. Breaking the regular expression down a bit:
^ matches the start of the line.
//.* matches // followed by zero or more characters.
\(//.*\)\{0,1\} matches the preceding regular expression 0 or 1 times (i.e., optionally)
$ matches the end of the line.
I am merging a number of text files on a linux server but the lines in some differ slightly and I need to unify them.
For example some files will have line like
id='1244' group='american' name='fred',american
Other files will be like
id='2345' name='frank', english
finally others will be like
id='7897' group='' name='maria',scottish
what I need to do is, if group='' or group is not in the string at all I need to add it somewhere before the comma setting it to the text after the comma so in the 2nd example above the line would become:
id='2345' name='frank' group='english',english
and the same in the last example which would become
id='7897' name='maria' group='scottish',scottish
This is going into a bash script. I can't actually delete the line and add to the end of the file as it relates to the following line.
I've used the following:
sed -i.bak 's#group=""##' file
which deletes the group="" string so the lines will either contain group='something' or wont contain it at all and that works
Then I tried to add the group if it doesn't exist using the following:
sed -i.bak '/group/! s#,(.*$)#group="\1",\1#' file
but that throws up the error
sed: -e expression #1, char 38: invalid reference \1 on `s' command's RHS
EDIT by Ed Morton to create a single sample input file and expected output:
Sample Input:
id='1244' group='american' name='fred',american
foo
id='2345' name='frank', english
bar
id='7897' group='' name='maria',scottish
Expected Output:
id='1244' group='american' name='fred',american
foo
id='2345' name='frank' group='english',english
bar
id='7897' name='maria' group='scottish',scottish
sed -r "
/group=''/ s/// # group is empty, remove it
/group=/! s/,[[:blank:]]*(.+)/ group='\\1',\\1/ # group is missing, add it
" file
id='1244' group='american' name='fred',american
foo
id='2345' name='frank' group='english',english
bar
id='7897' name='maria' group='scottish',scottish
The foo and bar lines are untouched because the s/// command did not match a comma followed by characters.
something like
sed '
/^[^,]*group[^,]*,/ ! {
s/, *\(.*\)/ group='\''\1'\'', \1/
}
/^[^,]*group='\'\''/ {
s/group='\'\''\([^,]*\), *\(.*\)/group='\''\2'\''\1, \2/
}
'
This GNU awk may help:
awk -v sq="'" '
BEGIN{RS="[ ,\n]+"; FS="="; found=0}
$1=="group"{
if($2==sq sq)
{next}
else
{found=1}
}
NF>1{
printf "%s=%s ",$1,$2
}
NF==1{
if(!found)
{printf "group=%s",$1}
print ","$1
found=0
}
' file
The script relies on the record separator RS which is set to get all key='value' pairs.
If the key group isn't found or is empty, it is printed when reaching a record with only one field.
Note that the variable sq holds the single quote character and is used to detect empty group field.
Sed can be pretty ugly. And your data format appears to be somewhat inconsistent. This MIGHT work for you:
$ sed -e "/group='[a-z]/b e" -e "s/group='' *//" -e "s/,\([a-z]*\)$/ group='\1', /" -e ':e' input.txt
Broken out for easier reading, here's what we're doing:
/group='[a-z]/b e - If the line contains a valid group, branch to the end.
s/group='' *// - Remove any empty group,
s/,\([a-z]*\)$/ group='\1', / - add a new group based on your specs
:e - branch label for the first command.
And then the default action is to print the line.
I really don't like manipulating data this way. It's prone to error, and you'll be further ahead reading this data into something that accurately stores its data structure, then prints the data according to a new structure. A more robust solution would likely be tied directly to whatever is producing or consuming this data, and would not sit in the middle like this.
Is it possible to merge multiple blocks/lines into a "single" line?
So basically if the next line starts with the same "#Msg" tag then append it to the previous line. (Hard to explain, but my example speaks for itself) (The blocks are separated by a new/blank line)
My input file looks like this:
#Msg,00000
#Msg,00001
#Msg,00002
#Msg,00003
#Msg,00004
#Msg,00005
#Msg,00006
#Msg,00007
#Msg,00008
#Msg,00009
#Msg,00010
#Msg,00011
Output should be like this:
#Msg,00000
#Msg,00001 #Msg,00002
#Msg,00003 #Msg,00004
#Msg,00005
#Msg,00006 #Msg,00007 #Msg,00008
#Msg,00009
#Msg,00010 #Msg,00011
Any advice is very welcome.
This would be pretty easy to do in Perl:
perl -00 -ple 'tr/\n/ /'
-e CODE specifies the program.
-p wraps a read/write line loop around it (by default it reads from STDIN, but you can also specify one or more filenames on the command line).
-00 specifies that the input "lines" are actually paragraphs.
-l has two effects: Incoming line terminators are automatically stripped from lines, and outgoing lines get line terminators added to them (and because we used -00 (paragraph mode), our line terminator is actually \n\n).
To recap:
We read the input one paragraph at a time. For each paragraph, we remove any trailing newlines. We then translate every newline to a space. Finally we output the transformed paragraph, followed by \n\n.
No point in trying to produce a shorter code than is possible with Perl!
Collect lines from the input file in list group until a blank line appears. Then output the contents of group, empty it and start again. When end-of-file is encountered output whatever is in group, if it is non-empty.
group = []
with open('vollschauer.txt') as vollschauer:
for line in vollschauer:
line = line.rstrip()
if line:
group.append(line)
else:
if group:
print (' '.join(group))
print()
group = []
if group:
print (' '.join(group))
group = []
$ awk -v RS= -v ORS='\n\n' '{$1=$1}1' file
#Msg,00000
#Msg,00001 #Msg,00002
#Msg,00003 #Msg,00004
#Msg,00005
#Msg,00006 #Msg,00007 #Msg,00008
#Msg,00009
#Msg,00010 #Msg,00011
If you insist on using sed, this should do the trick:
sed -r ':a; N; /^(#[^,]+,).*\n\1/! { P; D }; s/\n/ /; ba' file
It takes different tags into account. Such tags won't be grouped together (that's what I understood is the desired behavior):
$ cat file
#Msg,00000
#Msg,00001
#Hello,00002
#Hello,00003
#What,00004
#What,00005
$ sed -r ':a; N; /^(#[^,]+,).*\n\1/! { P; D }; s/\n/ /; ba' file
#Msg,00000 #Msg,00001
#Hello,00002
#Hello,00003
#What,00004 #What,00005
Note that this solution uses GNU sed.
This might work for you (GNU sed):
sed ':a;N;/^$/M!s/\n/ /;ta' file
Gather up lines, replacing each newline by a space until an empty line.
N.B. The use of the M flag on the repexp /^$/ which matches an empty line on a pattern space containing multiple lines.
Why outputs of these two commands differ?
cat config.xml|perl -ne 'print $1,"\n" if /([0-9\.]+):161/'
cat config.xml|perl -ne "print $1,"\n" if /([0-9\.]+):161/"
First works as expected printing out matched group while seconds prints whole line.
I see two main things wrong with your command.
First off, double quotes allow shell interpolation, and $1 will be taken for a shell variable and replaced. Since it unlikely exists, it will be replaced with an empty string. So instead of print $1, you get print, which is shorthand for print $_, and is probably why the entire line prints.
Second, you have unescaped double quotes inside your command, so you are in fact passing three strings to Perl:
print ,
\n
if /(....)/
As for why or how this works with your shell, I don't know, since I do not have access to your OS, nor know which one it is. In Windows, I get a Perl bareword warning for n (Unquoted string "n" may clash with future reserved word at -e line 1.) which means that the \n is interpreted as a string. Now, here's the tricky part. What we get is this:
print , \n if /.../
Which means that \n is no longer an argument to print, it is a statement that comes after print and it is in void context, so it gets ignored. We can see this by this warning (which I had to fake in my shell):
Useless use of single ref constructor in void context at -e line 1.
(Note that you do not get these warnings as you do not use warnings -- the -w switch)
So what we are left with is
print if /.../
Which is exactly the code for the behaviour you described: It prints the whole line when a match is found.
What you can do to visualize the problem in your shell is add the -MO=Deparse switch to your one-liner, as shown here:
C:\perl>perl -MO=Deparse -ne"print ,"\n" if /a/"
LINE: while (defined($_ = <ARGV>)) {
print($_), \'n' if /a/;
}
-e syntax OK
Now we can clearly see that the print statement is separated from the newline, and that the newline is a reference to a string.
Solution:
However, your code has other problems, and if done right you can avoid all the shell difficulties. First, you have a UUOC (Useless Use of Cat). A file argument can be given to perl when using the -n switch on the command line. Secondly, you do not need to use variables for this, you can simply print the return value of your regex:
perl -nlwe 'print for /(...)/' config.xml
The -l switch will handle newlines for you, and in this case add newline to the print. The for is necessary to avoid printing empty matches.
Inside double quote, some stuffs are substituted ($variable, `command`, ..). While inside single quote, they are remained as is.
$ echo "$HOME"
/home/falsetru
$ echo '$HOME'
$HOME
$ echo "`echo 1`"
1
$ echo '`echo 1`'
`echo 1`
Nested quotes:
$ echo ""hello""
hello
$ echo '"hello"'
"hello"
$ echo "\"hello\""
"hello"
Escape double quotes, $ to get same result:
cat config.xml | perl -ne "print \$1,\"\n\" if /([0-9\.]+):161/"
Two things:
Nested quotes.
Variables expand differently.
The first command has one string that happens to contain some double quotes. The variable is not expanded.
The second command has two strings with an unquoted \n in between. The variable is expanded.
Let's say $1 contains "blah"
The first passes this string to perl:
print $1,"\n" if /([0-9\.]+):161/
the second, this:
print blah,\n if /([0-9\.]+):161/
I have the file with the following format:
(Type 1 data:1) B B (Type 1 data:2) B B B
(Type 1 data:3) B ..
Now I want to reformat this file so that it looks like:
(Type 1 data:1) B B (Type 1 data:2) B B B (Type 1 data:3) B
...
My approach was to use perl regex in command line,
cat file | perl -pe 's/\n(B)/ $1/smg'
My reasoning was to replace the new line character with space.
but it doesn't seem to work. can you please help me? Thanks
The -p reads a line at a time, so there is nothing after the "\n" to match with.
perl -pe 'chomp; $_ = ($_ =~ /Type/) ? "\n".$_ : " ".$_'
this does almost what you want but puts one extra newline at the beginning and loses the final newline.
If the only place that ( shows up is at the beginning of where you want your lines to start, then you could use this command.
perl -l -0x28 -ne's/\n/ /g;print"($_"if$_' < file
-l causes print to add \n on the end of each line it prints.
-0x28 causes it to split on ( instead of on \n.
-n causes it to loop on the input. Basically it adds while(<>){chomp $_; to the beginning, and } at the end of what ever is in -e.
s/\n/ /g
print "($_" if $_ The if $_ part just stops it from printing an extra line at the beginning.
It's a little more involved as -n and -p fit best for processing one line at a time while your requirement is to combine several lines, which means you'd have to maintain state for a while.
So just read the entire file in memory and apply the regex like this:
perl -lwe ^
"local $/; local $_ = <>; print join q( ), split /\n/ for m/^\(Type [^(]*/gsm"
Feed your file to this prog on STDIN using input redirection (<).
Note this syntax is for the Windows command line. For Bash, use single quotes to quote the script.