Sed regex - include original matching

I found this solution using sed:

sed -n '/^[0-9]\{4\}\.[01][0-9]\.[0123][0-9]\./,${:a;N;$!ba;{s/\([0-9]\{4\}\.[01][0-9]\.[0123][0-9]\.\)/--------------\n\1/g;p}}'

The disadvantage is that the date has to be matched twice. Maybe there's another (better) solution.
The output is exactly as you expect in your example.

In other words you want to insert the line ---------- before every line that contains a YYYY.MM.DD date followed by a space and a bunch of lowercase letters. There are several ways to do this. You can use the insert command (i):

sed -e '/^[0-9][0-9][0-9][0-9]\.[0-9][0-9]\.[0-9][0-9] [a-z][a-z]*$/ i \
----------'

Or you can replace the empty string at the beginning of the line by a newline.

sed -e '/^[0-9][0-9][0-9][0-9]\.[0-9][0-9]\.[0-9][0-9] [a-z][a-z]*$/ s/^/----------\
'

Or you can use & in the replacement text of an s command to stand for the matched pattern.

sed -e 's/^[0-9][0-9][0-9][0-9]\.[0-9][0-9]\.[0-9][0-9] [a-z][a-z]*$/----------\
&'

Some sed implementations allow you to write \n instead of backslash-newline in the replacement text, but on others \n prints \n or n.

You should use awk instead

awk ' /[0-9]{4}\.[0-9]{2}\.[0-9]{2}\. / { print "---------------------\n" $0 ; continue } /^/ { print $0 } ' <"INPUTFILE" >"OUTPUTFILE"

basically it works in 2 steps:

step1: /[0-9]{4}\.[0-9]{2}\.[0-9]{2}\. / { print "---------------------\n" $0 ; continue }

means: if it maches /4digits.2digits.2digits. / then print "---...--\n" followed by the matching line, and loop on the next line (= "continue").

step2: /^/ { print $0 }

means: if we didn't match the above, then for all other lines (ie, matching a beginning of line, so even an empty line gets matched), just print that line.