How to print only last column?
Try:
echo -e 'one two three\nfour five six\nseven eight nine' | awk '{print $NF}'
It's easier than you think.
$ echo one two three | awk '{print $NF}'
three
Try grep
(shorter/simpler, but 3x slower than awk
because of regex usage):
grep -o '\S\+$' <(echo -e '... seven eight nine')
Or ex
(even more slower, but it prints the whole buffer after finish, more useful when it needs to be sorted or edited in-place):
ex -s +'%s/^.*\s//g' -c'%p|q!' <(echo -e '... seven eight nine')
ex +'%norm $Bd0' -sc'%p|q!' infile
To change in-place, replace -sc'%p|q!'
with -scwq
.
Or bash
:
while read line; do arr=($line); echo ${arr[-1]}; done < someinput
Performance
Given the generated 1GB file via:
$ hexdump -C /dev/urandom | rev | head -c1G | pv > datafile
I've performed the parsing time stats (ran ~3x and took the lowest, tested on MBP OS X):
using
awk
:$ time awk '{print $NF}' datafile > /dev/null real 0m12.124s user 0m10.704s sys 0m0.709s
using
grep
:$ time grep -o '\S\+$' datafile > /dev/null real 0m36.731s user 0m36.244s sys 0m0.401s $ time grep -o '\S*$' datafile > /dev/null real 0m40.865s user 0m39.756s sys 0m0.415s
using
perl
:$ time perl -lane 'print $F[-1]' datafile > /dev/null real 0m48.292s user 0m47.601s sys 0m0.396s
using
rev
+cut
:$ time (rev|cut -d' ' -f1|rev) < datafile > /dev/null $ time rev datafile | cut -d' ' -f1 | rev > /dev/null real 1m10.342s user 1m19.940s sys 0m1.263s
using
ex
:$ time ex +'%norm $Bd0_' -sc'%p|q!' datafile > /dev/null real 3m47.332s user 3m42.037s sys 0m2.617s $ time ex +'%norm $Bd0' -sc'%p|q!' datafile > /dev/null real 4m1.527s user 3m44.219s sys 0m6.164s $ time ex +'%s/^.*\s//g' -sc'%p|q!' datafile > /dev/null real 4m16.717s user 4m5.334s sys 0m5.076s
using
bash
:$ time while read line; do arr=($line); echo ${arr[-1]}; done < datafile > /dev/null real 9m42.807s user 8m12.553s sys 1m1.955s