How do I use cut to separate by multiple whitespace?

If we use tr command along with squeeze option (-s flag ) to convert all multiple consecutive spaces to a single space and then perform cut operation with space as delimiter – we can access the required column carrying the numbers.

Refer to the code snipped bellow:

cat file | tr -s ' ' | cut -d ' ' -f 8

To answer your question literally:

sed 's/   */:/g' | cut -d : -f 5

or

awk -F '  +' '{print $5}'

But that won't do if the number in brackets reaches 10, etc. If you're only interested in the numbers, you could remove everything else.

sed 's/[^.0-9][^.0-9]*/:/g' | cut -d : -f 6

These commands will all print the last column of a space separated file:

• awk '{print $NF}' file

  in awk, NF is the number of fields and $NF is the last field.

• perl -lane 'print $F[$#F]' file

  -a splits the file on whitespace into the array @F, $#F is the number of elements in the array so $F[$#F] is the last element. The -n means read the file given on the command line and apply the script passed with -e to each line. -l just adds a newline character (\n) to each print statement.

• sed 's/.* //g'

  a simple regular expression that matches everything to the last space and deletes it, leaving only the last column.

• rev file | cut -d' ' -f 1 | rev

  rev reverses its output so the last field is the first, cut with delimiter space to print it and rev to reverse the text back to normal. This won' t work if you have consecutive whitespace.

Based on your input, I am guessing you don't actually want the last column but the penultimate one or the two last ones. In that case use these to print the last 2 (8.39 Mbits/sec):

awk '{print $(NF-1),$NF}' file 
perl -lane 'print "$F[$#F-1] $F[$#F]"' file 
sed 's/.* \(.* .*\)/\1/' file 
rev file | cut -d' ' -f 1,2 | rev

and these to print the penultimate (8.39):

awk '{print $(NF-1)}' file 
perl -lane 'print $F[$#F-1]' file 
sed 's/.* \(.*\) .*/\1/' file 
rev file | cut -d' ' -f 2 | rev