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

Answer 2

And for yet another option, I'd go with awk!

echo "foo bar <foo> bla 1 2 3.4" | awk '{ print $(NF-1), $NF; }'

This will split the input (I'm using STDIN here, but your input could easily be a file) on spaces, and then print out the last-but-one field, and then the last field. The $NF variables hold the number of fields found after exploding on spaces.

The benefit of this is that it doesn't matter if what precedes the last two fields changes, as long as you only ever want the last two it'll continue to work.

Answer 3

grep is the right tool for extracting.

using your example and your regex:

kent$  echo 'foo bar <foo> bla 1 2 3.4'|grep -o '[0-9][0-9]*[\ \t][0-9.]*[\ \t]*$'
2 3.4

Answer 4

I agree with @kent that this is well suited for grep -o. If you need to extract a group within a pattern, you can do it with a 2nd grep.

# To extract \1 from /xx([0-9]+)yy/
$ echo "aa678bb xx123yy xx4yy aa42 aa9bb" | grep -Eo 'xx[0-9]+yy' | grep -Eo '[0-9]+'
123
4

# To extract \1 from /a([0-9]+)b/
$ echo "aa678bb xx123yy xx4yy aa42 aa9bb" | grep -Eo 'a[0-9]+b' | grep -Eo '[0-9]+'
678
9

Answer 5

The cut command is designed for this exact situation. It will "cut" on any delimiter and then you can specify which chunks should be output.

For instance: echo "foo bar <foo> bla 1 2 3.4" | cut -d " " -f 6-7

Will result in output of: 2 3.4

-d sets the delimiter

-f selects the range of 'fields' to output, in this case, it's the 6th through 7th chunks of the original string. You can also specify the range as a list, such as 6,7.