Shell Programming and Scripting

This is an vi editor question. I do not know is this a right place to ask this question or not?
I have a file with the following contents,

Code:

10 11 
20 21
30 31

I want to copy first column that is 10,20,30 after second column, so that output will look like the following,

Code:

10 11 10
20 21 20
30 31 30

How can I do that in VI editor through command mode (i.e. ":")?

Try this:

Code:

%s/\(.*\) .*/& \1/

%s/\(.*\) .*/& \1/

Substitute

---------- Post updated at 12:56 PM ---------- Previous update was at 12:54 PM ----------

%s/\(.*\) .*/& \1/

s :substitute
.* means number of characters
\ means do not take other meaning
what does & mean?
Can you explain the command please?

Code:

%s/\(.*\) .*/& \1/

Explanation:

%s -> substitute in the whole file

The \( and \) pair are stored in a buffer. The first \( and \) pair will represent by \1, the second \( and \) is \2 and so on

\(.*\) -> The first pair is everything before the last space

.* -> (space followed by a dot and an asterix) the last space and everything thereafter

& -> represent the entire line here

& \1/ -> means substitute the line with entire line followed by a space and the first pair