How to print only last column?

Question

echo -e 'one two three\nfour five six\nseven eight nine'
one two three
four five six
seven eight nine

how can I do some "MAGIC" do get this output?:

three
six
nine

UPDATE: I don't need it in this specific way, I need a general solution so that no matter how many columns are in a row, e.g.: awk always displays the last column.

Answer 1

Try:

echo -e 'one two three\nfour five six\nseven eight nine' | awk '{print $NF}'

Answer 2

It's easier than you think.

$ echo one two three | awk '{print $NF}'
three

Answer 3

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
  

Answer 4

It can even be done only with 'bash', without 'sed', 'awk' or 'perl':

echo -e 'one two three\nfour five six\nseven eight nine' |
  while IFS=" " read -r -a line; do
    nb=${#line[@]}
    echo ${line[$((nb - 1))]}
  done

Answer 5

... | perl -lane 'print $F[-1]'

Answer 6

It can also be done using 'sed':

echo -e 'one two three\nfour five six\nseven eight nine' | sed -e 's/^.* \([^ ]*\)$/\1/'

Update:

or more simply:

echo -e 'one two three\nfour five six\nseven eight nine' | sed -e 's/^.* //'

Answer 7

Or using cut:

echo -e 'one two three\nfour five six\nseven eight nine' | cut -f 3 -d' '

although this does not satisfy the 'general solution' requirement. Using rev twice we can solve this as well:

echo -e 'one two three\nfour five six\nseven eight nine' | rev | cut -f 1 -d' ' | rev

Answer 8

In perl this can be done as follows:

#!/usr/bin/perl

#create a line of arbitrary data
$line = "1 2 3 4 5";

# splt the line into an array (we call the array 'array', for lolz)
@array = split(' ', $line);

# print the last element in the array, followed by a newline character;
print "$array[-1]\n";

output:

$ perl last.pl
5
$

You could also loop through a file, heres an example script I wrote to parse a file called budget.dat

example data in budget.dat:

Rent              500
Food              250
Car               300
Tax               100
Car Tax           120
Mag Subscription  15

(you can see i needed to capture only the "last" column, not column 2 only)

The script:

#!/usr/bin/perl
$budgetfile = "budget.dat";
open($bf, $budgetfile)
        or die "Could not open filename: $filename $!";


print "-" x 50, "\n";
while ( $row = <$bf> ) {
        chomp $row;
        @r = split (' ', $row);
        print "$row ";
        $subtotal += $r[-1];
        print "\t$subtotal\n";
}
print "-" x 50, "\n";
print "\t\t\t Total:\t$subtotal\n\n";

Answer 9

Using awk you can first check if there is at least one column.

echo | awk '{if (NF >= 1) print $NF}'

echo 1 2 3 | awk '{if (NF >= 1) print $NF}'