How to split a line by multiple characters?

Question

Output of w | head -1 can give 2 different results for my machine:

16:34:42 up 32 days,  3:02,  1 user,  load average: 0.14, 0.11, 0.07
14:10:55 up 32 days, 39 min, 1 user, load average: 0.07, 0.13, 0.09

I need to get load average numbers (e.g. 0.07, 0.13, 0.09). How can I split the line by load average: string so that I can always get the load average numbers for two cases?

Answer 1

You can also use awk:

$ w | head -1 | awk '{print $10,$11,$12}'
0.80, 0.84, 0.93

Or, if the number of fields is variable, use:

$ w | head -1 | awk '{print $(NF-2),$(NF-1),$NF}'
0.81, 0.82, 0.91

Or, the much more elegant (thanks @Letitzia):

$ w | head -1 | awk -F "load average: " '{print $2}'

Sed:

$ w | head -1 | sed 's/.*load average: *//' 

Perl:

$ w | head -1 | perl -pe 's/.*load average: *//' 

or

$ w | head -1 | perl -lne '/.*load average: *(.*)/ && print $1' 

You can actually use colrm here but note that it removes "columns" which are defined as "a single character in a line". So, while it works for the example you have provided (because of the extra space) it won't if the number of columns changes further. Here, it's deleting columns 1 through 51

$ w | head -1 | colrm 1 51

Answer 2

Normally I'd just use the ${parameter#word} bash parameter expansion. It expands $parameter, deleting word (which can be a pattern) from the start.

In your case, something like:

line=...
echo ${line#*load average: }

Making it a function:

get_load() {
    w | head -n 1 | { read -r line; echo ${line#*load average: }; }
}

Answer 3

If you want only last numbers you can use grep:

$ w | grep -Po 'load average: \K.*'
0.07, 0.13, 0.09

Answer 4

Another perl:

$ w | head -1 | perl -nle 'print +(split /load average:/)[-1]'
 0.42, 0.49, 0.63