How to sort on column for rows containing a certain word

Question

I want to sort on a certain column only for rows containing a certain word. I don't want to see rows not containing that word. For example I have this text file:

 sdf ggfds   7
 sdf sgs     5
 sdf dfgs    3
 foo dffg    2
 bar dffg    2
 sdf sddfg   4

I want to sort 3rd column for rows containing only "sdf" word (doesnt have to be in a first column)

and I want to see this output:

 sdf dfgs    3
 sdf sddfg   4
 sdf sgs     5
 sdf ggfds   7

Answer 1

Pipe your input to an external command:

:%!grep sdf | sort -n -k3

Details:

1. select the whole content using '%'
2. pipe it to an external command using '!'
3. grep onyl the lines containing 'sdf'
4. sort these lines numerically (-n) on the third field (-k3)

Answer 2

2 vim commands:

:v/sdf/d
:sort n /[^[:digit:]]*/

• first deletes all lines that do not contain 'sdf'
• second sorts numbers ignoring non-numbers

Answer 3

Maxim Kim has already given an excellent answer and I was going to add this in a comment, but it just got too complicated so I'll stick it in an answer:

You could simplify the pattern by using:

:v/sdf/d
sort n /\D*/

as \D is equivalent to [^[:digit:]] and is a lot less typing. For more information, see

:help \D

To match on the third field specifically, rather than just the first digit, use

:sort n /\(\S\+\s+\)\{2}/`

or

:sort n /\v(\S+\s+){2}/

See:

:help :sort
:help \S
:help \s
:help pattern.txt
:help \v

As an aside, some find it easier to remember :g!/sdf/d, which does the same as :v/sdf/d - :g! is the opposite of :g and is identical to :v.

:help :v
:help :g