I've tinkered with Bash prompts for a lot of years. That's where this problem originates, but it can be considered just as a thought experiment if you prefer. We have the directory we're in, for example: /Users/gorr/.bashprompt/really/deep/directory/structure/even/deeper (I keep my prompts in ~/.bashprompt/ , and a very long directory name is often a cause of breakage, so I keep one around to break them ...) I want to shorten the directory name to just the first letters: export newPWD=$(echo ${PWD} | sed -e "s@${HOME}@~@" -e 's@ \(/.\)[^/]*@\1@g') echo $newPWD ~/./r/d/d/s/e/d This does '~' replacement for $HOME and then substitutes the first letter of each directory for the full directory name. Here's the question: if the first letter of the directory name is a dot '.', can sed then capture one character more so that the output would become: ~/.b/r/d/d/s/e/d I think this would be pretty easy with Bash and a loop, but that's a lot of processing so I'd rather not go down that road. I suspect sed is capable of this, but I haven't delved deeply enough into the tool to even know where to start. This may in fact be a regex problem more than a sed problem - either way I'm kind of stumped. I'm open to simpler implementations using other (standard system) tools as well. -- Giles https://www.gilesorr.com/ gilesorr@gmail.com