Joining 3 lines to the line that matches a pattern

Question

I have a file in VI Editor like this:

I1 a b c d e f 
g h i j k l m     
o p q r s t u v     
w x y z     
I2 a b c d     
e f g h i j k l m     
n o p q r s t     
u v w x y z    
I3 a  b c d e     
f g h i j k l m n     
o p q r s t u v w x y z

I'm trying to merge the 3 line that are after the line starting with I(^I) like this:

I1 a b c d e f g h i j k l m n o p q r s t u v w x y z     
I2 a b c d e f g h i j k l m n o p q r s t u v w x y z    
I3 a b c d e f g h i j k l m n o p q r s t u v w x y z

I have googled to see if I can join the 3 lines after the line containing I1 (that is, the line beginning with I [^I]) in VI editor and found the Join command that joins the next line to the current line like :g/^I/norm Jx. But I would like to use this command for joining the next 3 lines to the current line.

It would be highly appreciated if any one can suggest me a method to do this via VI Editor or any scripting language.

Is the 3 lines a hard number? You want to join 3 lines but your example joins 4, 4, and 3 lines. — sebnukem, Oct 30 '14 at 16:14
In VI (or better yet VIM): Go to the top line, press shift + J. Done. — TLP, Oct 30 '14 at 16:15
Hi: My mistake in the last sample. It should have been 4 lines. Thank for the reminder. — user3641866, Oct 30 '14 at 16:16

score 4 · Answer 1 · edited May 23 '17 at 11:57

Here's one way you could do it using awk:

awk 'NR>1&&/^I[0-9]/{print ""}{printf "%s", $0}END{print ""}' file.txt

When the line number is greater than 1 and the line starts with "I" followed by a digit, use print "" to print a newline. Use printf to print the contents of every line. In the END block (thanks fedorqui), print a final newline.

Testing it out on your file:

$ awk 'NR>1&&/^I[0-9]/{print ""}{printf "%s", $0}END{print ""}' file.txt
I1 a b c d e fg h i j k l mo p q r s t u vw x y z
I2 a b c de f g h i j k l mn o p q r s tu v w x y z
I3 a b c d ef g h i j k l m no p q r s t u v w x y z

Note this misses a final end line. You can solve it by using `END {print ""}`. — fedorqui, Oct 30 '14 at 16:24

glenn jackman · Answer 2 · 2014-10-30T18:22:05.960

3

You're almost there:

:g/^I/norm 4J

:g takes a regex and a command
:norm 4J is an "ex" command that executes "normal mode" commands. i.e. what you'd type.

See :help :g and :help :norm

Based on Peter's comment: :g/^I/.,+3join or :g/^I/j4

edited Oct 30 '14 at 18:22

answered Oct 30 '14 at 16:20

glenn jackman

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Some explanation for VI novices such as myself would be appreciated :) – Tom Fenech Oct 30 '14 at 16:21
I think globally `:g` search for `/^I/` perform normal command `norm` and `4J` is a repeat command of shift+j (join lines). – TLP Oct 30 '14 at 16:24
This will fail on the last block of the given sample because it doesn't have enough lines. – romainl Oct 30 '14 at 16:55
1

Forget `:normal` and just use `:join` – Peter Rincker Oct 30 '14 at 17:12
2

I love the range but you don't need it in this case for `:join`. e.g `:g/^I/j4` – Peter Rincker Oct 30 '14 at 17:22
Ah, missed the version with {count} in the docs – glenn jackman Oct 30 '14 at 18:20

romainl · Answer 3 · 2014-10-30T16:53:20.397

2

If you are sure every block has the same number of lines, say 4:

:g/^I/norm 4J

If you can't be sure about the number of lines to join the problem becomes slightly more complex:

:g/^I/norm O    " separates every block with a blank line
:g//vipJ        " join each block into it's own line         
:g/^$/d         " removes every blank line

If you don't mind thinking a little bit outside the box, this command will work with any block size:

:%join|s/ I/\rI/g    " join the whole buffer into one single line
                     " then substitute every ' I' with `\rI`

edited Oct 30 '14 at 16:53

answered Oct 30 '14 at 16:47

romainl

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+1 I like the `:join`/`:s` approach myself. If you don't mind the last block not being joined you can do `:g/^I/,//-j` to join arbitrary number of lines between blocks – Peter Rincker Oct 30 '14 at 17:15
For joining all the lines, it could be necessary to rotect against 'ignorecase': `%join|s/ \CI/\rI/g` – mMontu Oct 30 '14 at 17:29

score 2 · Answer 4 · answered Oct 30 '14 at 16:52

2

tr -d "\n" <filename | sed 's/ \+/ /g;s/ I/\nI/g'

Output:

I1 a b c d e f g h i j k l m o p q r s t u v w x y z
I2 a b c d e f g h i j k l m n o p q r s t u v w x y z
I3 a b c d e f g h i j k l m n o p q r s t u v w x y z

answered Oct 30 '14 at 16:52

Cyrus

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fedorqui · Answer 5 · 2014-10-30T16:23:29.190

1

This awk makes it:

awk '/^I[0-9]/ {if (f) print f; f=""} {f=sprintf("%s%s", (f?f FS:""), $0)} END {print f}' file

It keeps adding the lines into a variable f. When a line starting with I + digit is found, it prints it.

for the given input returns:

I1 a b c d e f g h i j k l m o p q r s t u v w x y z
I2 a b c d e f g h i j k l m n o p q r s t u v w x y z
I3 a  b c d e f g h i j k l m n o p q r s t u v w x y z

edited Oct 30 '14 at 16:23

answered Oct 30 '14 at 16:16

fedorqui

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Thanks fedorqui. This worked too. I have both the methods. Thanks for the immediate reply and help. – user3641866 Oct 30 '14 at 16:28

score 1 · Answer 6 · answered Oct 30 '14 at 16:19

1

Perlish answer;

#!/usr/bin/perl

use strict;
use warnings;

while ( <DATA> ) {
   chomp;
   s/\s+/ /g;
   if ( m/^I/ ) { print "\n" };
   print;
}

__DATA__
I1 a b c d e f 
g h i j k l m     
o p q r s t u v     
w x y z     
I2 a b c d     
e f g h i j k l m     
n o p q r s t     
u v w x y z    
I3 a  b c d e     
f g h i j k l m n     
o p q r s t u v w x y z

answered Oct 30 '14 at 16:19

Sobrique

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Thanks Sobrique. Built my script based on yours and some modifications. – user3641866 Oct 30 '14 at 21:37

score 1 · Answer 7 · answered Oct 30 '14 at 16:23

A solution in pure bash:

#! /bin/bash

start="I"
cur=
while read line ; do
    if test "${line:0:1}" = "$start" ; then
        test "$cur" = "" || { echo "$cur" ; cur= ;}
    fi
    cur+="$line"
done << EOT
I1 a b c d e f 
g h i j k l m     
o p q r s t u v     
w x y z     
I2 a b c d     
e f g h i j k l m     
n o p q r s t     
u v w x y z    
I3 a  b c d e     
f g h i j k l m n     
o p q r s t u v w x y z 
EOT

echo "$cur"

mMontu · Answer 8 · 2014-10-30T18:25:04.930

On Vim, in addition to the ex commands pointed by glen and romainl, you could record a macro:

qm/^I<enter>vnJq

Then repeat it twice:

2@m

Explanation

qm - start recording a macro on a given register (mon this case)
/^Ienter - search for regex ^I
v - start visual selection
n - select text before the next occurrence of the regex (this step and the previous could be replaced for V3j if the number of lines is fixed)
J - join selected lines
q - stop recording the macro

Joining 3 lines to the line that matches a pattern

8 Answers8