mjhjmtu

I am new to regex and sed, and am trying to create what I thought would be a straightforward regex: I want to remove word-final letter if it's an 'o'.

• Input string: Hello Hello


• Expected output: Hell Hell

The good news: I can remove the 'o' when it is at the end of the string:

$ echo 'Hello Hello' |sed 's/(.*)o/1/g'
Hello Hell
$ echo 'Hello Hello' |sed 's/(.*)o$/1/g'
Hello Hell

The bad news: I cannot remove it from words earlier in the string. I have tried this with all the anchor symbols I can think of. The result is that none of the word-final 'o's is removed:

$ echo 'Hello Hello' |sed 's/(.*)ob/1/g'
Hello Hello
$ echo 'Hello Hello' |sed 's/(.*)o>/1/g'
Hello Hello
$ echo 'Hello Hello' |sed 's/(.*)oW/1/g'
Hello Hello
$ echo 'Hello Hello' |sed 's/(.*)os/1/g'
Hello Hello

Can you please help me regain my sanity by telling me what I'm doing wrong?

Update: I get the dictinct impression that my machine produces different results than some other people's. I am using the terminal window on my Macbook. If anyone can shed some light on this, please tell me.

echo 'Hello Hello' | sed 's/o$//'

seems more useful to me than your

echo 'Hello Hello' | sed 's/(.*)o$/1/g'

In your question is says that the output of

echo 'Hello Hello' | sed 's/(.*)ob/1/g'

was Hello Hello but for me it is Hello Hell. You can correct that to

echo 'Hello Hello' | sed 's/([^o]*)ob/1/g'

but

echo 'Hello Hello' | sed 's/ob//g'

seems better to me.

Removing the o at the end of words is removing a o between a word character and a non-word character (or the EOL) so:

sed -r 's/(w)o(W|$)/12/g'

I'm wondering if somehow space isn't your word delimeter. Try something like the following:

$ echo hello hello | sed -e 's/o / /g;s/o$//'
hell hell

The problem with this example is that you'll also have to do the same for . and , and any other word delimeter. Match o followed by another specific char with like o[ .,]. For some reason this doesn't work for EOL$, so add another search string with ;. Example:

$ echo hello hello, hello. toot hello | sed -e 's/o([ .,])/1/g;s/o$//'
hell hell, hell. toot hell
$ echo $SHELL
/bin/bash
$ sed --version
sed (GNU sed) 4.4
$ set | grep IFS
IFS=$' tn'

I have tried this with all the anchor symbols I can think of.

It's not the anchors, but the fact that you have a greedy match with the asterisk. The (.*)o expression matches as long a string as it can, so
it will eat everything up to the last o. It might match earlier o's too.

But then, capturing something and then returning it back is useless, you could just remove the (.*) and the 1 completely.

So, these would (at least in GNU sed) remove o's at the end of words:

sed 's/o>//g' 
sed 's/ob//g' 

This, of course only at the end of string:

sed 's/o$//g' 

And this will remove an o, along with a following non-word-character (e.g. the space after Hello):

sed 's/oW//g' 

If your sed doesn't support </> or b, you'll have to do something else. This would match o followed by a non-alphanumeric character, or the end of line:

$ echo "jello, jello" | sed -E -e 's/o([^[:alnum:]]|$)/1/g'
jell, jell

This works e.g. in the sed that comes with OS X/macOS.

Perl regexes support adding a question mark to * or + to make them non-greedy. Then they would match the shortest possible string:

echo "jello, jello" | perl -pe 's/(.*?)o/$1/g'
jell, jell