Grep – How to Run Grep on a Single Column

awkgrep

I want to grep the output of my ls -l command:

-rw-r--r--   1 root root       1866 Feb 14 07:47 rahmu.file
-rw-r--r--   1 rahmu user     95653 Feb 14 07:47 foo.file
-rw-r--r--   1 rahmu user   1073822 Feb 14 21:01 bar.file

I want to run grep rahmu on column $3 only, so the output of my grep command should look like this:

-rw-r--r--   1 rahmu user     95653 Feb 14 07:47 foo.file
-rw-r--r--   1 rahmu user   1073822 Feb 14 21:01 bar.file

What's the simplest way to do it? The answer must be portable across many Unices, preferably focusing on Linux and Solaris.

NB: I'm not looking for a way to find all the files belonging to a given user. This example was only given to make my question clearer.

Best Answer

One more time awk saves the day!

Here's a straightforward way to do it, with a relatively simple syntax:

ls -l | awk '{if ($3 == "rahmu") print $0;}'

or even simpler: (Thanks to Peter.O in the comments)

ls -l | awk '$3 == "rahmu"'

Related Solutions

Sed / Awk – Find string exactly n characters long

awk 'length($1) == 12 { print $1 }' file

The program is pretty self documenting and avoids the regex hammer.

awk -v f=1 '$f ~ /^[[:alnum:]]{12}$/ { print $f }' file

Or swing away with the above if you only want to consider first fields (fields being delimited by blanks) that consist only of alphanumeric characters.

With awk implementations that don't support the {x,y} interval regular expressions, you can change it to:

awk -v f=1 'length($f) == 12 && $f !~ /[^[:alnum:]]/ { print $f }' file

Linux – Split a File into Rows Based on Column Values

The solution for potentially unsorted input file:

sort -k1,1 -k2,2n file | awk '$1=="chr4" && $2>2 && $2<8'

The output:

chr4    3    C    435
chr4    4    A    223
chr4    5    T    400
chr4    6    G    300
chr4    7    G    340

If the input file is sorted it's enough to use:

awk '$1=="chr4" && $2>2 && $2<8' file

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