Replace whole line containing a string using Sed

Try tail:

tail -n +2 "$FILE"

-n x: Just print the last x lines. tail -n 5 would give you the last 5 lines of the input. The + sign kind of inverts the argument and make tail print anything but the first x-1 lines. tail -n +1 would print the whole file, tail -n +2 everything but the first line, etc.

GNU tail is much faster than sed. tail is also available on BSD and the -n +2 flag is consistent across both tools. Check the FreeBSD or OS X man pages for more.

The BSD version can be much slower than sed, though. I wonder how they managed that; tail should just read a file line by line while sed does pretty complex operations involving interpreting a script, applying regular expressions and the like.

Note: You may be tempted to use

# THIS WILL GIVE YOU AN EMPTY FILE!
tail -n +2 "$FILE" > "$FILE"

but this will give you an empty file. The reason is that the redirection (>) happens before tail is invoked by the shell:

1. Shell truncates file $FILE
2. Shell creates a new process for tail
3. Shell redirects stdout of the tail process to $FILE
4. tail reads from the now empty $FILE

If you want to remove the first line inside the file, you should use:

tail -n +2 "$FILE" > "$FILE.tmp" && mv "$FILE.tmp" "$FILE"

The && will make sure that the file doesn't get overwritten when there is a problem.

Bit of awk should do it?

awk '{s+=$1} END {print s}' mydatafile

Note: some versions of awk have some odd behaviours if you are going to be adding anything exceeding 2^31 (2147483647). See comments for more background. One suggestion is to use printf rather than print:

awk '{s+=$1} END {printf "%.0f", s}' mydatafile