Skip the first 6 lines/rows in a text file with awk

Question

How can I skip the first 6 lines/rows in a text file (input.txt) and process the rest with awk? The format of my awk script (program.awk) is:

BEGIN {
} 

{ 
process here
} 

END {

}

My text file is like this:

0
3
5
0.1 4.3
2.0 1.5
1.5 3.0
0.3 3.3
1.5 2.1
.
.
.

I want to process the file starting from:

0.3 3.3
1.5 2.1
.
.
.

Answer 1

Use either of the two patterns:

NR>6 { this_code_is_active }

or this:

NR<=6 { next }
{ this_code_is_active }

Use FNR instead of NR if you have many files as arguments to awk and want to skip 6 lines in every file.

Answer 2

Try:

awk 'FNR > 6 { #process here }' file

Answer 3

You may also skip an arbitrary number of lines at the beginning or the end of the file using head or tail programs.

For your concrete question,

tail input.txt -n+7 | program.awk

will do, provided your program.awk file is executable. Otherwise, you may use

tail input.txt -n+7 | awk -f program.awk

This way, you will spare a comparison for each line and you don't need to change the logic of your AWK code.

tail will start streaming text starting at the seventh line, skipping the six first lines.

This will not be a huge deal in performance, especially if text process is simple thanks to caching. However, for long files and repeated use in cloud environment may save some cost.

Answer 4

Another one:

$ cat print_from.awk

#print 20 lines from line 6 (from your file: 0.3 3.3)
BEGIN{n=0}
/^0.3 3.3.*/ {n=NR; m=n+20}
{
  if (n > 1 && NR < m) {print $1} 
}

execute with:

awk -f print_from.awk your-text-file

Answer 5

for pkg in $(pip list | awk 'FNR > 2 { print $1 }'); do pip install --upgrade $pkg; done