Separate columns in one line using Bash script

Question

I am using bash scripting in Linux. I have a variable that looks like this:

a="1-1-1 1-1-2 1-1-3 1-1-4"

I want to separate columns with values equal to or greater than 1-1-3 using AWK.

My output should be:

1-1-3 1-1-4

Can someone help me with this?

How do you define "greater than"? What about a="2-1-1 1-1-2 1-1-3 1-1-4" ? — P.P
– P.P, Commented Jan 23, 2016 at 8:29
I think at first , column one should be compared then two and three, regarding to your example, output will be 2-1-1 1-1-3 1-1-4. if our data is single column I can use echo "$a | awk '($0 >= 1-1-3) {print $0}' ... but now I have input in one line — user3891747
– user3891747, Commented Jan 23, 2016 at 8:41

Casimir et Hippolyte · Accepted Answer · 2016-01-23 12:12:40Z

1

You can do it without a loop, if you use the space as record separator:

echo $a | awk -v RS=' ' -v ORS=' ' '{$1=$1} $1>="1-1-3"'

answered Jan 23, 2016 at 12:12

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Comments

P.P · Accepted Answer · 2016-01-23 08:54:37Z

1

You can loop over the line and compare:

echo $a | awk -v ORS=' ' '{for(i=1;i<=NF;i++) if($i >= "1-1-3") print $i;}'

answered Jan 23, 2016 at 8:54

P.P

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glenn jackman · Accepted Answer · 2016-01-23 16:04:04Z

1

sort's -V option gives you "version" sorting that might be helpful:

$ a="2-0-0 1-1-1 1-1-2 1-1-3 1-1-4"
$ tr ' ' '\n' <<<"$a" | sort -V
1-1-1
1-1-2
1-1-3
1-1-4
2-0-0

answered Jan 23, 2016 at 16:04

glenn jackman

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