sort but keep header line in the at the top?

Question

I am getting output from a program that first produces one line that is a bunch of column headers, and then a bunch of lines of data. I want to cut various columns of this output and view it sorted according to various columns. Without the headers, the cutting and sorting is easily accomplished via the -k option to sort along with cut or awk to view a subset of the columns. However, this method of sorting mixes the column headers in with the rest of the lines of output. Is there an easy way to keep the headers at the top?

Answer 1

Stealing Andy's idea and making it a function so it's easier to use:

# print the header (the first line of input)
# and then run the specified command on the body (the rest of the input)
# use it in a pipeline, e.g. ps | body grep somepattern
body() {
    IFS= read -r header
    printf '%s\n' "$header"
    "$@"
}

Now I can do:

$ ps -o pid,comm | body sort -k2
  PID COMMAND
24759 bash
31276 bash
31032 less
31177 less
31020 man
31167 man
...

$ ps -o pid,comm | body grep less
  PID COMMAND
31032 less
31177 less

Answer 2

You can keep the header at the top like this with bash:

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

Or do it with perl:

command | perl -e 'print scalar (<>); print sort { ... } <>'

Answer 3

Hackish but effective: prepend 0 to all header lines and 1 to all other lines before sorting. Strip the first character after sorting.

… |
awk '{print (NR <= 2 ? "0 " : "1 ") $0}' |
sort -k 1 -k… |
cut -b 3-

Answer 4

Here's some magic perl line noise that you can pipe your output through to sort everything but keep the first line at the top: perl -e 'print scalar <>, sort <>;'

Answer 5

I found a nice awk version that works nicely in scripts:

awk 'NR == 1; NR > 1 {print $0 | "sort -n"}'