Deleting lines matching regex along with 2 previous lines

Question

I have a file where I need to remove the 2 lines before the regex foo.

bad 1
foo
good 1
good 2
good 3
bad 2
bad 3
foo
good 4
good 5
good 6
bad 4
bad 5
foo
good 7
bad 6
bad 7
foo
good 8
good 9
good 10
bad 8
bad 9
foo
good11

I can do it easily with sed :

casper_mint@casper-mint-dell /tmp $ cat dddd | sed '/bad/ , /foo/d'
good 1
good 2
good 3
good 4
good 5
good 6
good 7
good 8
good 9
good 10
good11

How can I do this in perl?

Answer 1

In a one liner:

print -ne 'push @b, $_; @b = () if /^foo$/; print shift @b if @b > 2; END {print @b};' file

Or demonstrated in a full script:

use strict;
use warnings;

my @b;

while (<DATA>) {
    push @b, $_;
    @b = () if /^foo/;
    print shift @b if @b > 2;
}

END {print @b};

__DATA__
bad 1
foo
good 1
good 2
good 3
bad 2
bad 3
foo
good 4
good 5
good 6
bad 4
bad 5
foo
good 7
bad 6
bad 7
foo
good 8
good 9
good 10
bad 8
bad 9
foo
good11

Both Output:

good 1
good 2
good 3
good 4
good 5
good 6
good 7
good 8
good 9
good 10
good11

Answer 2

In perl, you could make use of the range operator:

perl -ne 'print unless /bad/ .. /foo/' filename

would give the same output as sed.

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You may also want to look at s2p.

Related: How can I pull out lines between two patterns that are themselves on different lines?