Is there a way to ignore header lines in a UNIX sort?

Question

(head -n 2 <file> && tail -n +3 <file> | sort) > newfile

The parentheses create a subshell, wrapping up the stdout so you can pipe it or redirect it as if it had come from a single command.

If you don't mind using awk, you can take advantage of awk's built-in pipe abilities

eg.

extract_data | awk 'NR<3{print $0;next}{print $0| "sort -r"}' 

This prints the first two lines verbatim and pipes the rest through sort.

Note that this has the very specific advantage of being able to selectively sort parts of a piped input. all the other methods suggested will only sort plain files which can be read multiple times. This works on anything.

Here is a version that works on piped data:

(read -r; printf "%s\n" "$REPLY"; sort)

If your header has multiple lines:

(for i in $(seq $HEADER_ROWS); do read -r; printf "%s\n" "$REPLY"; done; sort)

This solution is from here

In simple cases, sed can do the job elegantly:

    your_script | (sed -u 1q; sort)

or equivalently,

    cat your_data | (sed -u 1q; sort)

The key is in the 1q -- print first line (header) and quit (leaving the rest of the input to sort).

For the example given, 2q will do the trick.

The -u switch (unbuffered) is required for those seds (notably, GNU's) that would otherwise read the input in chunks, thereby consuming data that you want to go through sort instead.

You can use tail -n +3 <file> | sort ... (tail will output the file contents from the 3rd line).

head -2 <your_file> && nawk 'NR>2' <your_file> | sort

example:

> cat temp
10
8
1
2
3
4
5
> head -2 temp && nawk 'NR>2' temp | sort -r
10
8
5
4
3
2
1