sed: print only matching group

Question

I want to grab the last two numbers (one int, one float; followed by optional whitespace) and print only them.

Example:

foo bar  bla 1 2 3         


        

Answer 1

Match the whole line, so add a .* at the beginning of your regex. This causes the entire line to be replaced with the contents of the group

echo "foo bar  bla 1 2 3.4" |
 sed -n  's/.*\([0-9][0-9]*[\ \t][0-9.]*[ \t]*$\)/\1/p'
2 3.4