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sed Exact Match when Dot is present

 
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# 1  
Old 03-18-2015
Linux sed Exact Match when Dot is present
I am trying to replace exact word from my text. I know using the angled brackets does the trick. But it is not working when there is a dot in the text.

Code:
echo "Bottle BottleWater Bottle Can" | sed 's/\<Bottle\>//g'
 BottleWater  Can

But if my data has a dot or hash in it, it replaces all the occurrences of Bottle.
Code:
echo "Bottle Bottle.Water Bottle Can" | sed 's/\<Bottle\>//g'
 .Water  Can

Code:
echo "Bottle Bottle#Water Bottle Can" | sed 's/\<Bottle\>//g'
 #Water  Can

As you can see it is replacing all occurances of "Bottle" when there is a dot present.
The input data is not in my control. So I cant change or escape the characters.
Is there a way I can replace the exact match using sed, when a dot is present in the data?
I have to use Sed, since my code is generating a lot of sed text using -e options so I can run all of that in one go.

Any help would be much appreciated.

Thanks.
# 2  
Old 03-19-2015
Lose the g -

Code:
 echo "Bottle Bottle#Water Bottle Can" | sed 's/\<Bottle\>//'
 Bottle#Water Bottle Can

This will replace the FIRST occurrence of the word. One replacement only per line of text.
# 3  
Old 03-19-2015
Right, I am aware of that Jim. But I want only the exact words to be replaced. Not all.
It should not remove the Bottle from "Bottle.Water".
Is there a way to tell sed to treat dots like underscores while searching?
# 4  
Old 03-19-2015
Some versions of sed have options to use EREs instead of BREs which could shorten this considerably. Since you haven't told us what OS you're using, I'll give you the long, portable form (assuming your word delimiters are start of line, end of line, and space characters):
Code:
printf '%s\n' "$string" | sed -e 's/^Bottle / /' -e 's/ Bottle$/ /' -e 's/^Bottle$//' -e 's/ Bottle /  /g'

Last edited by Don Cragun; 03-19-2015 at 11:43 AM.. Reason: Change: -e '/s/^Bottle$//' to -e 's/^Bottle$//'
# 5  
Old 03-19-2015
Or try awk:
Code:
awk -v s="Bottle" '{for(i=1; i<=NF; i++) if($i==s) $i=""}1'

---
Quote:
Originally Posted by grep_me
I am trying to replace exact word from my text. I know using the angled brackets does the trick. But it is not working when there is a dot in the text.
The escaped angle brackets (\< and \>) indicate a (left/right) word boundary in some versions of sed (alphanumeric characters plus _ (underscore)).
Since a dot is not a word character, "Bottle" will get replaced..
Last edited by Scrutinizer; 03-19-2015 at 02:11 AM..
# 6  
Old 03-19-2015
Quote:
Originally Posted by Scrutinizer
Or try awk:
Code:
awk -v s="Bottle" '{for(i=1; i<=NF; i++) if($i==s) $i=""}1'

... ... ...
Note, however, adjacent spaces in the input will be coalesced to a single space by the above command. For instance.
Code:
echo 'Bottle rocket.  Bottled juices.  Bottled beer.' | awk -v s="Bottle" '{for(i=1; i<=NF; i++) if($i==s) $i=""}1'

produces:
Code:
 rocket. Bottled juices. Bottled beer.

removing one space after the first two periods in the input string.

It isn't clear from the skimpy description of the desired results in this thread whether or not this is allowable. But, the request was for sed solutions only...
# 7  
Old 03-19-2015
Yes, I was suggesting an alternative to sed. The space squeezing could be avoided by using
Code:
awk -F '[ ]' ...

but that only works if there are no TABs..


--
In the case of your sed solution TABs can be factored in by using [[:blank:]] instead of space:
Code:
sed -e 's/^Bottle\([[:blank:]]\)/\1/' -e 's/\([[:blank:]]\)Bottle$/\1/' -e 's/^Bottle$//' -e 's/\([[:blank:]]\)Bottle\([[:blank:]]\)/\1\2/g'


--
There is a spurious leading / -e '/s/^Bottle$//'

--
But then there is still a problem with a text like foo Bottle Bottle bar
Last edited by Scrutinizer; 03-19-2015 at 06:30 AM..
This User Gave Thanks to Scrutinizer For This Post:
Don Cragun
 
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