I want to grab the last two numbers (one int, one float; followed by optional whitespace) and print only them.
Example:
foo bar <foo> bla 1 2 3.4
Should print:
2 3.4
So far, I have the following:
sed -n 's/\([0-9][0-9]*[\ \t][0-9.]*[\ \t]*$\)/replacement/p'
will give me
foo bar <foo> bla 1 replacement
However, if I try to replace it with group 1, the whole line is printed.
sed -n 's/\([0-9][0-9]*[\ \t][0-9.]*[\ \t]*$\)/\1/p'
How can I print only the section of the line that matches the regex in the group?
Answer
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 <foo> bla 1 2 3.4" |
sed -n 's/.*\([0-9][0-9]*[\ \t][0-9.]*[ \t]*$\)/\1/p'
2 3.4