SED command to delete empty lines till the first occurrence of sentence - regex

Question

My input file will be
[emptyline]
[emptyline]
aaa
bbb
[emptyline]
cc
dd
Here [emptyline] indicates blanklines.
And I need an SED command to change this into
aaa
bbb
[emptyline]
cc
dd
That is, I need to delete all the blank lines at the top alone.
I need only SED command since i need to use that in bash script.
Additional info its MAC OSx

Answer 1

You can do it with branching in sed:
sed '/^ *$/d; :a; n; ba' file
A more efficient solution would be to use a range expression, see user2719058's answer for how to do this.
It is even more efficient if you can reduce the need for sed, see gniourf_gniourf's answer for alternatives.
This can be expressed in awk elegantly like this:
awk 'NF {f=1} f' file
Output in both cases:
aaa
bbb
cc
dd
Explanation
Both alternatives work by looking for the first non-empty line.
With sed the pattern /^ *$/d will delete all empty lines in the beginning of the file. What follows is a loop that prints the rest of the file.
awk will update NF for every line, when the line is empty NF is zero. This is exploited for setting the print-flag (f).

Answer 2

If the lines are really empty (no whitespace), I would suggest
sed -n '/./,$p', otherwise sed -n $'/[^ \t]/,$p'. (The $'..' syntax makes bash expand the \t, so you don't need a sed that understands it.)

Answer 3

One funny possibility:
{ sed -n '/./{p;q}' && cat; } < file
And it's really efficient too! (try to benchmark it against the other methods). If you might have some spaces in your first lines, you could do:
{ sed -n '/[^[:space:]]/{p;q}' && cat; } < file
sed does nothing until it reads a character; at this point it prints out the line and exits. Then cat outputs the whole thing; so since there's no more sed filtering, the data flows much faster through cat!
The same with grep:
{ grep -v -m 1 '^$' && cat; } < file
or discarding leading lines with possible spaces:
{ grep -v -m 1 '^[[:space:]]*$' && cat; } < file

Answer 4

A simple one is: sed '1,/^$/d' file
It will delete starting at line 1 up to the last blank line prior to the actual content of file; preserving the other blank lines as desired by OP.

Answer 5

Here is another way of deleting all blank lines at file start using pure BASH way without involving any external utility like awk/sed:
[[ "$(<file)" =~ ^[[:space:]]+(.*)$ ]] && echo "${BASH_REMATCH[1]}"
aaa
bbb
cc
dd

Answer 6

sed -n "H;$ {x;s/^\n*//p;}"
delete all first \n ant take into account that 1st line is maybe not empty (1,/^$/ does not work in this case)