Rearrange text file by prepending data rows with header row

Question

I have a file whose content looks like this:

2009 150  0  0  0     0000
75.316   0.0400390625  0.00007        0.00000 0.8980
76.216   0.0400390625  0.00007        1.00000 0.9046
77.217   0.0400390625  0.00009        2.00000 0.9113
78.341   0.0400390625  0.00010        3.00000 0.9183
2009 150  2  0  0     0000
75.316   0.0400390625  0.00007        0.00000 0.8980
76.216   0.0400390625  0.00007        1.00000 0.9046
77.217   0.0400390625  0.00009        2.00000 0.9113
78.341   0.0400390625  0.00010        3.00000 0.9183
79.616   0.0400390625  0.00013        4.00000 0.9255

I would like to:

1. Find all lines that start with keyword 2009.
2. Pre-pend those lines to all following ones until another occurrence of line that starts with 2009 is found, and so on until EOF.

In essence I am looking for an output like this:

2009 150  0  0  0     0000 75.316   0.0400390625  0.00007        0.00000 0.8980
2009 150  0  0  0     0000 76.216   0.0400390625  0.00007        1.00000 0.9046
2009 150  0  0  0     0000 77.217   0.0400390625  0.00009        2.00000 0.9113
2009 150  0  0  0     0000 78.341   0.0400390625  0.00010        3.00000 0.9183
2009 150  2  0  0     0000 75.316   0.0400390625  0.00007        0.00000 0.8980
2009 150  2  0  0     0000 76.216   0.0400390625  0.00007        1.00000 0.9046

.......

I have been scratching my head on this for quite a while and cannot come up with a solution. Help would be appreciated. I know how to extract the text between keywords using flags but I am not sure if that is the right direction.

Answer 1

You can use awk for this:

awk '/^2009/{a=$0;next}{print a" "$0} ' file.txt

This will only prepend a space to any lines before the first match of 2009, you can set a default string to prepend like this:

awk 'BEGIN{a="My default prepend string";}/^2009/{a=$0;next}{print a" "$0} ' file.txt

Answer 2

I'd do this in Perl:

perl -lne 'if(/^2009/){$n=$_; next} print "$n $_"' file

or, more concisely

perl -lne '/^2009/ ? ($n=$_) : print "$n $_"' file

The idea is to save the current line ($_ in Perl) as $n if it begins with 2009 and if not, print the current line along with the current value of $n.

Answer 3

Using sed:

sed -n '
  /^2009/ { h }
  /^2009/ !{ G; s/^\(.*\)\n\(.*\)$/\2 \1/p }
  ' in_file

Explanation as requested:

-n - makes sed not print anything unless we tell it to.

/^2009/ { h } - when we reach a line beginning with 2009, put it in the hold buffer.

/^2009/ !{...} - the pattern inside the {...} is gets applied to every line that doesn't start with 2009.

G; s/^\(.*\)\n\(.*\)$/\2 \1/p - G copies the hold buffer and appends it the line just read into the pattern space. There are now two lines in the pattern space which we need to swap the order of and combine before printing. The replace pattern with back-references does just this.