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

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 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.