I want to print the second-to-last column or field in awk
. The number of fields is the NF
variable. I know that I should be able to use $NF
, but I'm not sure how it can be used.
And this does not seem to work:
awk ' { print ( $NF-- ) } '
Small addition to Chris Kannon' accepted answer: only print if there actually is a second last column.
(
echo | awk 'NF && NF-1 { print ( $(NF-1) ) }'
echo 1 | awk 'NF && NF-1 { print ( $(NF-1) ) }'
echo 1 2 | awk 'NF && NF-1 { print ( $(NF-1) ) }'
echo 1 2 3 | awk 'NF && NF-1 { print ( $(NF-1) ) }'
)
You weren't far from the result! This does it:
awk '{NF--; print $NF}' file
This decrements the number of fields in one, so that $NF
contains the former penultimate.
Let's generate some numbers and print them on groups of 5:
$ seq 12 | xargs -n5
1 2 3 4 5
6 7 8 9 10
11 12
Let's print the penultimate on each line:
$ seq 12 | xargs -n5 | awk '{NF--; print $NF}'
4
9
11
Perl solution similar to Chris Kannon's awk solution:
perl -lane 'print $F[$#F-1]' file
These command-line options are used:
n
loop around every line of the input file, do not automatically print every line
l
removes newlines before processing, and adds them back in afterwards
a
autosplit mode – split input lines into the @F
array. Defaults to splitting on whitespace
e
execute the perl code
The @F
autosplit array starts at index [0] while awk fields start with $1.$#F
is the number of elements in @F
NF
is the last field index,$NF
is the value of the last field - glenn jackman