Tell a regex expression to skip the beginning of a line before matching

Since you clarified what you wanted in a comment to Gnouc's answer, here's a solution:

sed 's|[^ /]*/|--|g'

$ echo '45mb ./aaaa/bbbb/cccc/dddd' | sed 's|[^ /]*/|--|g'
45mb --------dddd

This will break if you have a trailing slash, or if the filepath you're passing to it contains any spaces. It would be pretty easy to script something a little more watertight, but it would involve more than a single line.

Here's one solution using capture groups:

sed -e 's|\([^ ]* \).*/\(.*\)|\1\2|'

I'm using |s as the separators because then I don't have to bother escaping the forward-slashes (but I would have to escape any |s in the pattern). AFAIK sed can have pretty much any character as a separator.

The first capture group \([^ ]* \) matches 'any number of any character except a blank space, followed by a blank space'. The .*/ matches 'any number of any characters, followed by a forward-slash', and the second group \(.*\) captures 'any number of any characters'.

Sed's regular expressions (and most regular expressions) are greedy by default, so .*/ will match the longest string that matches its pattern.

$ echo '45mb ./aaaa/bbbb/cccc/dddd' | sed 's|\([^ ]* \).*/\(.*\)|\1\2|'
45mb dddd

However, this will break if there is a trailing slash:

$ echo '45mb ./aaaa/bbbb/cccc/dddd/' | sed 's|\([^ ]* \).*/\(.*\)|\1\2|'
45mb 

This version will work even with a trailing slash, but will break if you have more than one:

sed -e 's|\([^ ]* \).*/\(.\)|\1\2|'

$ echo '45mb ./aaaa/bbbb/cccc/dddd/' | sed 's|\([^ ]* \).*/\(.\)|\1\2|'
45mb dddd/
$ echo '45mb ./aaaa/bbbb/cccc/dddd//' | sed 's|\([^ ]* \).*/\(.\)|\1\2|'
45mb /

Here is solution using regex with perl:

$ echo '45mb ./aaaa/bbbb/cccc/dddd' | perl -ne 'print $1 if /(\.\/.*\/)/'
./aaaa/bbbb/cccc/    

And substitution:

$ echo '45mb ./aaaa/bbbb/cccc/dddd' | perl -ple 's/(\.\/.*\/)/--/'
45mb --dddd