Reverse four length of letters with sed in unix

Question

How can I reverse a four length of letters with sed?

For example:

the year was 1815.

Reverse to:

the raey was 5181.

This is my attempt:

cat filename | sed's/\([a-z]*\) *\([a-z]*\)/\2, \1/'

But it does not work as I intended.

Answer 1

not sure it is possible to do it with GNU sed for all cases. If _ doesn't occur immediately before/after four letter words, you can use

sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'

\b is word boundary, word definition being any alphabet or digit or underscore character. So \b will ensure to match only whole words not part of words

$ echo 'the year was 1815.' | sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
the raey was 5181.
$ echo 'two time five three six good' | sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
two emit evif three six doog

$ # but won't work if there are underscores around the words
$ echo '_good food' | sed -E 's/\b([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])\b/\4\3\2\1/gi'
_good doof

tool with lookaround support would work for all cases

$ echo '_good food' | perl -pe 's/(?<![a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])([a-z0-9])(?!=[a-z0-9])/$4$3$2$1/gi'
_doog doof

(?<![a-z0-9]) and (?!=[a-z0-9]) are negative lookbehind and negative lookahead respectively

Can be shortened to

perl -pe 's/(?<![a-z0-9])[a-z0-9]{4}(?!=[a-z0-9])/reverse $&/gie'

which uses the e modifier to place Perl code in substitution section. This form is suitable to easily change length of words to be reversed

Answer 2

Possible shortest sed solution even if a four length of letters contains _s.

sed -r 's/\<(.)(.)(.)(.)\>/\4\3\2\1/g'