How can I post-process a capture group while preserving others in sed?

user_13632026 0
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I have a string with 3 capture groups and would like to preserve the first and third but perform a substitution on the second. How do I express this in sed?

Concretely, I have an input string like:

top-level.subpath.one.subpath.two.subpath.forty-five

And I want to preserve the part before the first ., shorten the middle part to the first letter of every word, and preserve the part after the last .. The result should look like:

top-level.s.o.s.t.s.forty-five

For preserving the capture groups, I have:

sed -r 's/([^.]*)(.*)(\..*)/\1...\3/'

which gets me:

top-level....forty-five

For converting something like .subpath.one.subpath.two.subpath to only initials, I have:

sed -r 's/(\.[^.])[^\.]*/\1/g'

which gets me:

.s.o.s.t.s

I'd like to essentially apply that second sed expression to capture group 2. Is there some way I can chain sed substitutions to perform that second substitution on only the second capture group while retaining the first and third?

sed bash regex
5 Answers

A simple awk solution that will work with any version of awk including MacOS:

s='top-level.subpath.one.subpath.two.subpath.forty-five'
awk 'BEGIN{FS=OFS="."} {for(i=2;i<NF;++i) $i=substr($i,1,1)}1' <<< "$s"

top-level.s.o.s.t.s.forty-five

This awk command uses . as input and output field separator. We loop through field position 2 to last-1 and replace value of each field with the first character of that field. In the end we print full record.

A BSD sed solution to do the same:

sed -E -e ':x' -e 's/(.+\..)[^.]+\./\1./; tx' <<< "$s"

top-level.s.o.s.t.s.forty-five
user_5482250

You can use

sed -E ':a; s/^(.*\.[^.])[^.]+(\.)/\1\2/; ta' file > newfile         # GNU sed
sed -E -e :a -e 's/^(.*\.[^.])[^.]+(\.)/\1\2/' -e ta file > newfile  # FreeBSD sed

See the online demo. Details:

  • -E - enables POSIX ERE syntax (+ is now a one or more quantifier, (...) is parsed as a grouping construct)
  • :a - sets an a label
  • s/^(.*\.[^.])[^.]+(\.)/\1\2/ - finds zero or more chars, a . and then any single char other than a . (capturing this into Group 1), then one or more chars other than a ., and then matches and captures into Group 2 a dot char, the match is replaced with concatenated Group 1 + Group 2 values
  • ta - goes to the a label upon successful replacement.
user_38329700

This might work for you (GNU sed):

sed -E ':a;s/(\..*)\B.(.*\.)/\1\2/;ta' file

Capture the first and last periods and hollow out the middle removing any side-by-side word characters.

Ameliorating @anubhava's sed answer:

sed -E 's/(\..)[^.]+\./\1./g;s//\1./g' file

Using the global flag and repeating the same substitution provides a 2 command solution.

user_9674920

To answer your question, it's not at all possible to run commands on the string/group captured by a single sub expression.

I would use this sed to achieve your task:

sed -E 's/(\.[^.])[^.]+\./\1./g; s/(\.[^.])[^.]+\./\1./g'
  • The two s commands are identical, but must be run twice to modify adjacent fields, as the pattern begins and ends with a (literal) dot to avoid matching the first and last field.
  • FWIW, this was consistently faster than sed -E -e :goto -e 's/(\.[^.])[^.]+\./\1./; t goto' in benchmarks - by 15% with GNU sed, and 25% with busybox sed.

There's also this method using GNU sed:

sed -E 's/([^.])[^.]*\./\1./2g'
  • Ng starts global replacement at the Nth match.
  • It's a documented feature in GNU sed, however POSIX says that combining a number with g produces undefined behaviour.
  • It could be useful if you needed to keep more than the first field.
user_139196680

With GNU awk for gensub():

$ awk 'BEGIN{FS=OFS="."} {beg=$1; end=$NF; $0=gensub(/([^.])[^.]+/,"\\1","g"); $1=beg; $NF=end} 1' file
top-level.s.o.s.t.s.forty-five
user_17450010
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