Shell: delete every second match against a regex i

问题:

Say I have come up with a regex matching a piece of data; the regex contains 2 sed groups (sub-expressions enclosed in ( and )). Also say that this regex is duplicated 9 times to match a whole line. The problem I am facing is how to delete (in an elegant way) every second match against the regex.

回答1:

Let's say you have the following string and want to remove the occurrences of bar:

foo bar foo bar foo bar

You can use the following sed command, note the option g which makes the substitution happen as many times as possible:

sed -r 's/([a-z]+) ([a-z]+)/\1/g' <<< 'foo bar foo bar foo bar'

Output: foo foo foo.

However this would not work with a string where the number of words is not even. I would make the second capturing group optional using the * quantifier to make the above commmand even work with such strings:

sed -r 's/([a-z]+) ([a-z]+)*/\1/g' <<< 'foo bar foo bar foo bar foo'

Output: foo foo foo foo.