bash - Print only parts that match regex -

echo "a b _c d _e f" | sed 's/[ ]*_[a-z]\+//g'

the result a b d f.

now, how can turn around, , print _c _e, while assuming nothing rest of line?

if question "how can print substrings match specific regular expression using sed?" hard achieve (and not obvious solution).

grep more helpful in case:

$> echo "a b _c d _e f" | grep -o -p "(\ *_[a-z]+)"  _c  _e

and finally

$> echo `echo "a b _c d _e f" | grep -o -p "(\ *_[a-z]+)"` _c _e

OSX