[ACCEPTED]-How to sort on column for rows containing a certain word-vim
Pipe your input to an external command:
:%!grep sdf | sort -n -k3
Details:
- select the whole content using '%'
- pipe it to an external command using '!'
- grep onyl the lines containing 'sdf'
- sort these lines numerically (-n) on the third field (-k3)
0
2 vim commands:
:v/sdf/d
:sort n /[^[:digit:]]*/
- first deletes all lines that do not contain 'sdf'
- second sorts numbers ignoring non-numbers
0
Maxim Kim has already given an excellent 11 answer and I was going to add this in a 10 comment, but it just got too complicated 9 so I'll stick it in an answer:
You could 8 simplify the pattern by using:
:v/sdf/d
sort n /\D*/
as \D is equivalent 7 to [^[:digit:]] and is a lot less typing. For more 6 information, see
:help \D
To match on the third field 5 specifically, rather than just the first 4 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 3 easier to remember :g!/sdf/d, which does the same 2 as :v/sdf/d - :g! is the opposite of :g and is identical 1 to :v.
:help :v
:help :g
Sort by 2nd column by selecting it in visual 11 mode (e.g. Control+v), then run:
!sort
or to sort by 3rd 10 column
sort -k 3
or
:sort /.*\%3v/
Alternatively select the lines you 9 wish to sort using the Shift+V command. Then enter
!sort -k 3n
or 8 use the below code to tell Vim to skip the 7 first two words in every line and sort on 6 whatever follows:
:%sort /^\S\+\s\+\S\+\s\+/
or i.e. sort by 8th line:
:sort /.*\%55v/
The 5 'virtual' specification is the absolute 4 number of column , which treats spaces + tabs 3 as single character (shortly, it doesn't 2 count tabs as eight spaces),
so to sort by 1 last column:
:%sort /\<\S\+\>$/ r
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