How do I print the last 5 fields in awk?

Question

I have 10 fields and I want to start from field 5 to field 10 and ignore the first 5 fields. How can I use NF in awk to do that?

f1 f2 f3 f4 f5 f6 f7 f8 f9 f10
c1 c2 c3 c4 c5 c6 c7 c8 c9 c10

I want to show only:

f6 f7 f8 f9 f10
c6 c7 c8 c9 c10

Solution 1

You need to loop through the fields:

bash-4.3$ awk '{for(i=6;i<=NF;i++) printf $i" "; print ""}' input_file.txt 
f6 f7 f8 f9 f10 
c6 c7 c8 c9 c10 

Or you can make fields equal to Null string:

bash-4.3$ awk '{for(i=1;i<=5;i++) $i="";print}' input_file.txt 
     f6 f7 f8 f9 f10
     c6 c7 c8 c9 c10

Or use substring of the whole line , to print all characters from where field 6 begins (credit to https://stackoverflow.com/a/12900372/3701431):

bash-4.3$ awk '{print substr($0,index($0,$6))}' input_file.txt 
f6 f7 f8 f9 f10
c6 c7 c8 c9 c10

or simply use cut command:

bash-4.3$ cut -d " " -f6-10  input_file.txt 
f6 f7 f8 f9 f10
c6 c7 c8 c9 c10

Python can do that too:

bash-4.3$ python -c 'import sys;fields=[" ".join(line.strip().split()[5:]) for line in sys.stdin];print "\n".join(fields)' < input_file.txt 
f6 f7 f8 f9 f10
c6 c7 c8 c9 c10

or alternatively:

$ python -c "import sys;print '\n'.join(map(lambda x:' '.join(x.split()[5:]),sys.stdin.readlines()))" < input_file.txt
f6 f7 f8 f9 f10
c6 c7 c8 c9 c10

Or with Ruby:

bash-4.3$ ruby -ne 'print $_.split()[5..10].join(" ");print "\n"' < input_file.txt 
f6 f7 f8 f9 f10
c6 c7 c8 c9 c10

Bash + xargs can do it too, although a bit more convoluted:

bash-4.3$ cat input_file.txt | xargs -L 1 bash -c 'arr=($@);for i in $(seq 5 10);do printf "%s " ${arr[$i]} ; done; echo' sh
f6 f7 f8 f9 f10  
c6 c7 c8 c9 c10 

Solution 2

Just process the fields of interest. That will be the last field -4, the last field -3, until the actual last field.

Reading from a file with this content (file.txt):

f1 f2 f3 f4 f5 f6 f7 f8 f9 f10
c1 c2 c3 c4 c5 c6 c7 c8 c9 c10

Run awk on the file as described below. The line with the $ sign is the command. The last two lines are the output.

$ awk '{print $(NF-4)" "$(NF-3)" "$(NF-2)" "$(NF-1)" "$NF}' file.txt
f6 f7 f8 f9 f10
c6 c7 c8 c9 c10

Note: As pointed out by Cyrus in the comment, I removed the bash script and left just the print statement to make it simpler and faster.

Comments

[muru]

stackoverflow.com/questions/13446255/…, stackoverflow.com/questions/16400819/…

[Walter A]

You ask for awk, so cut -f6- file for fixed 10 fields, or counting from the end with rev file | cut -f1-5 | rev may work but don't count as an answer.

[Cyrus]

or awk '{print $(NF-4)" "$(NF-3)" "$(NF-2)" "$(NF-1)" "$NF}' file.txt

[Apologician]

Thanks ! I'll add the shorter version to the answer. Leaving out the cat pipe.

[Barb Hammond]

I don't think the solution with index() will work if some of the other fields may be equal to or contain $6.

[Sergiy Kolodyazhnyy]

yes, this will work only in this specific case

[Walter A]

printf $i" " will fail when the field has a %.