Extracting Last column using VI

Question

I have a csv file which contains some 1000 fields with values, the headers are something like below:

v1,v2,v3,v4,v5....v1000

I want to extract the last column i.e. v1000 and its values.

I tried %s/,[^,]*$// , but this turns out to be exact opposite of what i expected, Is there any way to invert this expression in VI ?

I know it can be done using awk as awk -F "," '{print $NF}' myfile.csv, but i want to make it happen in VI with regular expression,Please also note that i have VI and don't have VIM and working on UNIX, so i can't do visual mode trick as well.

Many thanks in advance, Any help is much appreciated.

Answer 1

Don't you just want

%s/.*,\s*//

.*, is match everything unto the last comma and the \s* is there to remove whitespace if its there.

Answer 2

You already accepted answer, btw you can still use awk or other nice UNIX tools within VI or VIM. Technique below calls manipulating the contents of a buffer through an external command :!{cmd}

As a demo, let's rearrange the records in CSV file with sort command:

first,last,email
john,smith,john@example.com
jane,doe,jane@example.com

:2,$!sort -t',' -k2

-k2 flag will sort the records by second field.

Extract last column with awk as easy as:

:%!awk -F "," '{print $NF}'

Answer 3

Dont forget cut!

:%!cut -d , -f 6

Where 6 is the number of the last field.

Or if you don't want to count the number of fields:

:%!rev | cut -d , -f 1 | rev