How can I output only captured groups with sed?

Question

Is there a way to tell sed to output only captured groups?

For example, given the input:

This is a sample 123 text and some 987 numbers 

And pattern:

/([\d]+)/ 

Could I get only 123 and 987 output in the way formatted by back references?

Answer 1

The key to getting this to work is to tell sed to exclude what you don't want to be output as well as specifying what you do want.

string='This is a sample 123 text and some 987 numbers' echo "$string" | sed -rn 's/[^[:digit:]]*([[:digit:]]+)[^[:digit:]]+([[:digit:]]+)[^[:digit:]]*/\1 \2/p' 

This says:

• don't default to printing each line (-n)
• exclude zero or more non-digits
• include one or more digits
• exclude one or more non-digits
• include one or more digits
• exclude zero or more non-digits
• print the substitution (p)

In general, in sed you capture groups using parentheses and output what you capture using a back reference:

echo "foobarbaz" | sed 's/^foo\(.*\)baz$/\1/' 

will output "bar". If you use -r (-E for OS X) for extended regex, you don't need to escape the parentheses:

echo "foobarbaz" | sed -r 's/^foo(.*)baz$/\1/' 

There can be up to 9 capture groups and their back references. The back references are numbered in the order the groups appear, but they can be used in any order and can be repeated:

echo "foobarbaz" | sed -r 's/^foo(.*)b(.)z$/\2 \1 \2/' 

outputs "a bar a".

If you have GNU grep (it may also work in BSD, including OS X):

echo "$string" | grep -Po '\d+' 

or variations such as:

echo "$string" | grep -Po '(?<=\D )(\d+)' 

The -P option enables Perl Compatible Regular Expressions. See man 3 pcrepattern or man 3 pcresyntax.