Print only the last three columns from file

Question

I have this file on Linux:

 14:00:01.071 5255  604177
4 14:00:01.074 4608  1415742
 14:00:01.074 18398  1122001
2 14:00:01.074 11723  155575
5 14:00:01.075 4695  885808

Desired Output:

 14:00:01.071 5255  604177
 14:00:01.074 4608  1415742
 14:00:01.074 18398 1122001
 14:00:01.074 11723 155575
 14:00:01.075 4695  885808

Command Used:

gawk '{ print $NF-1, $NF}' filename

But it prints only the last two columns.

Answer 1

It is quite incorrect, $NF-1 would print the value of the last column subtracted by one. You need to group the command within (..), since NF is a variable

gawk 'NF>=3{ print $(NF-2), $(NF-1), $NF}' file

The part NF>=3 is to safely validate you are printing the last 3 fields on lines that is guaranteed to have at least 3 fields.

Answer 2

Assuming

1. The delimiter is a space
2. You want all but the first column, which may be blank, but will have a delimiter
3. There's only one delimiter between columns

then you can use cut -d' ' -f2- to print, from a space-separated values file, the contents from the second column to the last one.

Answer 3

$ awk '{ print $(NF-2), $(NF-1), $NF}' file1
14:00:01.071 5255 604177
14:00:01.074 4608 1415742
14:00:01.074 18398 1122001
14:00:01.074 11723 155575
14:00:01.075 4695 885808

Answer 4

With GNU grep, you can do:

grep -Eo '\S+\s+\S+\s+\S+\s*$'

Which would preserve the spacing in the input. Note that lines that contain fewer than 3 fields are skipped.

POSIXly, the equivalent would be:

S='[^[:space:]]\{1,\}' s='[[:space:]]\{1,\}'
sed -n "s/$S$s$S$s${S}[[:space:]]*$/\\
&/;s/.*\n//p"

Answer 5

I tested with below command and it worked fine

 sed "s/^[0-9]\{1,\}//g" filename| sed -r "s/^\s+//g"

output

14:00:01.071 5255  604177
14:00:01.074 4608  1415742
14:00:01.074 18398  1122001
14:00:01.074 11723  155575
14:00:01.075 4695  885808

Answer 6

awk '{print $(NF-2),$(NF-1),$(NF)}' file.txt

Where NF is the number of fields in your file.

Answer 7

Since the question is tagged perl

$ perl -lane 'print "@F[-3..-1]"' ip.txt
14:00:01.071 5255 604177
14:00:01.074 4608 1415742
14:00:01.074 18398 1122001
14:00:01.074 11723 155575
14:00:01.075 4695 885808

• -l strip newlines from input lines, and add it to end of string when print is used
• -a autosplit input based on whitespaces, accessible via @F array
• print "@F[-3..-1]" print the desired fields, the default delimiter when array is specified within double quotes is single space
  • -1 refers to last field, -2 refers to second last field, etc
  • .. is range operator

Use perl -lane 'print "@F[-3..-1]" if $#F > 2' if you do not want lines with less than 3 fields to be displayed. $#F gives the last index of array, i.e. size of array minus 1