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Full Discussion: Update a field using awk and keep the formatting.
Top Forums Shell Programming and Scripting Update a field using awk and keep the formatting. Post 302910527 by Don Cragun on Friday 25th of July 2014 05:39:29 AM
Old 07-25-2014
Anytime you reset a numbered field (such as $2=x or sub(regex, replacement, $n)) for n not equal to 0), $0 is recalculated. For the general case, you could try something like:
Code:
(	printf "  2 4 6\n"
	printf "1\t3\t7\t\n"
	printf "\t xyz\t 2\t 3 with more fields\n"
) | awk '
{	n = split($0, delims, /[^[:space:]]*/)
	$2+=3
	$3-=1
	for(i = 1; i <= n; i++)
		printf("%s%s", delims[i], $i)
	print ""
}'

which produces the output:
Code:
  2 7 5
1	6	6	
	 xyz	 5	 2 with more fields

If you want to try this on a Solaris/SunOS system, change awk to /usr/xpg4/bin/awk, /usr/xpg6/bin/awk, or nawk.

The loop printing the modified lines and the following print command can be simplified to a single printf command if you know that there are always three fields.
Last edited by Don Cragun; 07-25-2014 at 06:48 AM.. Reason: Add note explaining why problem occurs...
This User Gave Thanks to Don Cragun For This Post:
Jotne
 

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LEARN ABOUT DEBIAN

bup-margin

bup-margin(1)						      General Commands Manual						     bup-margin(1)

NAME

       bup-margin - figure out your deduplication safety margin

SYNOPSIS

       bup margin [options...]

DESCRIPTION

       bup margin  iterates  through  all  objects  in	your  bup repository, calculating the largest number of prefix bits shared between any two
       entries.  This number, n, identifies the longest subset of SHA-1 you could use and still encounter a collision between your object ids.

       For example, one system that was tested had a collection of 11 million objects (70 GB), and bup margin returned 45.  That  means  a  46-bit
       hash  would be sufficient to avoid all collisions among that set of objects; each object in that repository could be uniquely identified by
       its first 46 bits.

       The number of bits needed seems to increase by about 1 or 2 for every doubling of the number of objects.  Since SHA-1 hashes have 160 bits,
       that  leaves 115 bits of margin.  Of course, because SHA-1 hashes are essentially random, it's theoretically possible to use many more bits
       with far fewer objects.

       If you're paranoid about the possibility of SHA-1 collisions, you can monitor your repository by running bup margin occasionally to see	if
       you're getting dangerously close to 160 bits.

OPTIONS

       --predict
	      Guess  the offset into each index file where a particular object will appear, and report the maximum deviation of the correct answer
	      from the guess.  This is potentially useful for tuning an interpolation search algorithm.

       --ignore-midx
	      don't use .midx files, use only .idx files.  This is only really useful when used with --predict.

EXAMPLE

	      $ bup margin
	      Reading indexes: 100.00% (1612581/1612581), done.
	      40
	      40 matching prefix bits
	      1.94 bits per doubling
	      120 bits (61.86 doublings) remaining
	      4.19338e+18 times larger is possible

	      Everyone on earth could have 625878182 data sets
	      like yours, all in one repository, and we would
	      expect 1 object collision.

	      $ bup margin --predict
	      PackIdxList: using 1 index.
	      Reading indexes: 100.00% (1612581/1612581), done.
	      915 of 1612581 (0.057%)

SEE ALSO

       bup-midx(1), bup-save(1)

BUP

       Part of the bup(1) suite.

AUTHORS

       Avery Pennarun <apenwarr@gmail.com>.

Bup unknown-															     bup-margin(1)
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