[SOLVED] [BASH][SED] Find pattern and replace it's [pattern's] last character

czezz · 10-19-2020, 04:47 PM

Hi all,
Can anyone please give me a little help with sed here?

I have following example file:

Code:

cat test
blablalb A i B. E. lalalalal.
A. tralala,la A B i C.
B. elellfdew A. fww.
C. DiDweofjoie zalamki T.

Following grep will highlight the pattern that Im trying to change:

Code:

cat test | grep " i [A-Z][.]"   (mind the SPACE before "i")

So, eventually what Im trying to get is:

Code:

blablalb A i B, E. lalalalal.
A. tralala,la A B i C,
B. elellfdew A. fww.
C. DiDweofjoie zalamki T.

Normaly, if there was no pattern I would go for someting like this:

Code:

sed -e 's/ i [A-Z][.]/ i [A-Z][,]/g' test
sed -e '/ i [A-Z][.]/s//,/' test

Any ideas how to get it?

teckk · 10-19-2020, 04:54 PM

Code:

st="blablalb A i B. E. lalalalal.
A. tralala,la A B i C.
B. elellfdew A. fww.
C. DiDweofjoie zalamki T."

sed 's/\,/\./g' <<< "$st"
blablalb A i B. E. lalalalal.
A. tralala.la A B i C.
B. elellfdew A. fww.
C. DiDweofjoie zalamki T.

Or did I miss something.

Edit:You hightlighted/changed it while I was answering.

boughtonp · 10-19-2020, 05:16 PM

Don't think of it as "the last character", think of it as a pattern with two parts; you want to keep the first part and replace the second.

With Sed, you can use a capturing group around the first part, then reference it via its number in the replacement:

Code:

echo 'axbx' | sed -r 's/(a)x/\1y/'

With Perl, you have the additional option of using a lookbehind to match (but not replace) the first part:

Code:

echo 'axbx' | perl -pe 's/(?<=a)x/y/'

(The latter can be useful in situations where you already have capturing groups and don't want to renumber them.)

czezz · 10-19-2020, 05:28 PM

Hi guys, thank you for your replies.
@teckk - sorry about that. I thought it would be useful to highlight patterns as its a tricky one.
@boughtonp - I tried to use my pattern " i [A-Z][.]" but that doesnt work. Thanks anyway for reply

Note: this is to replace . with , ONLY when dot stands next to capital letter followed by " i " (mind the spaces)
eg.

Code:

i B. >>>  i B,

danielbmartin · 10-19-2020, 07:24 PM

With this InFile ...

Code:

blablalb A i B. E. lalalalal.
A. tralala,la A B i C.
B. elellfdew A. fww.
C. DiDweofjoie zalamki T.

... this sed ...

Code:

sed -r 's/(i [A-Z])(.)/\1\,/g' $InFile >$OutFile

... produced this OutFile ...

Code:

blablalb A i B, E. lalalalal.
A. tralala,la A B i C,
B. elellfdew A. fww.
C. DiDweofjoie zalamki T,

Daniel B. Martin

.

czezz · 10-20-2020, 07:44 AM

Thank you very much, Daniel

I did tiny modification to it as I need this SPACE before "i"

Code:

sed -r 's/( i [A-Z])(.)/\1\,/g'

pan64 · 10-20-2020, 07:56 AM

that is still not ok.
But first: you cannot use a regexp in the right side (replacement string), because there is no string to match. You need to specify the replacement string itself.
If you want to specify dot (.) you must use [.] (or \.), not (.) as regexp, and you need not use \ in the replacement string before the comma (but \1 is ok, because it has a special meaning).

MadeInGermany · 10-20-2020, 10:32 AM

With -r or -E option the left side is ERE (extended regular expression), where ( ) is a capture group.

Code:

sed -r 's/( i [A-Z])[.]/\1,/g' test

Without -r or -E option the left side is BRE (basic regular expression), where \( \) is a capture group.

Code:

sed 's/\( i [A-Z]\)[.]/\1,/g' test

The right side is the substitution string, where \1 is what the 1st capture group matched.

danielbmartin · 10-21-2020, 07:27 PM

OP asked for a sed solution and now has one. As a learning exercise I worked up an awk solution.

With this InFile ...

Code:

blablalb A i B. E. lalalalal.
A. tralala,la A B i C.
B. elellfdew A. fww.
C. DiDweofjoie zalamki T.

... this awk ...

Code:

awk '{if (n=match($0,/ i [A-Z]\./)) 
        {before=substr($0,n,5)
         after=substr($0,n,4)","
         sub(before,after)}
     }1' $InFile >$OutFile

... produced this OutFile ...

Code:

blablalb A i B, E. lalalalal.
A. tralala,la A B i C,
B. elellfdew A. fww.
C. DiDweofjoie zalamki T.

Corrections and improvements are always welcomed.

Daniel B. Martin

.

pan64 · 10-22-2020, 01:10 AM

you can do the same gensub in awk (quite similar to the sed solution)

Code:

awk ' { print gensub("( i [A-Z])\.", "\\1", "g" } '

(not tested, this is just the idea)

danielbmartin · 10-22-2020, 07:43 AM

Quote:

Originally Posted by pan64

you can do the same gensub in awk (quite similar to the sed solution)

Code:

awk ' { print gensub("( i [A-Z])\.", "\\1", "g" } '

(not tested, this is just the idea)

Excellent! With minor mods ...

Code:

awk '{print gensub("( i [A-Z])\\.", "\\1,", "g")}' $InFile >$OutFile

... this works nicely. {Tested.}

Love those one-line solutions!

Daniel B. Martin

.