Ubuntu – How to print the last 5 fields in awk

awkbashcommand linetext processing

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

Best Answer

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
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