how to delete the second character "." from the line

what I have is this ( but its remove the first "." from output

uname -r | sed s'/\./ /'

2 6.18-164.2.1.el5PAE

while I need the following output

2.6 18-164.2.1.el5PAE
  • 1
    Please also show us the input. I can extrapolate from the output and assume that your input is 2.6.18-164.2.1.el5PAE but including it would make the question clearer.
    – terdon
    Oct 5, 2014 at 16:48

2 Answers 2


Simply add N to the end of the command for it to match the Nth match, like this:

uname -r | sed 's/\./ /2'

What do you need it for though?

From the info page on sed:

The `s' command can be followed by zero or more of the following FLAGS:


Apply the replacement to _all_ matches to the REGEXP, not just the first.


Only replace the NUMBERth match of the REGEXP.
  • I need only the first number from uname -r ( 2.6 ) all other numbers are non relevant anyway thx on your good answer Oct 6, 2014 at 6:27

Here are a few ways to remove the second . from a line of a file (they will affect all lines of the file):

  1. sed. You already have what is probably the best way, but here's another:

    sed 's/\([^.]*\.[^.]*\)\./\1 /' file 

    This will look for the longest stretch of non-. ([^.]*), then a . (\.), then the next stretch of non-. and finally a . (\.). The parentheses capture the patterns so we can refer to it as \1. Therefore, the command above will just delete the second . and replace it with a space.

    If you have GNU sed (default on Linux), you can simplify to:

    sed -r 's/([^.]*\.[^.]*)\./\1 /' file 
  2. Perl

    perl -pe 's/([^.]*\.[^.]*)\./\1 /' file 


    perl -F'\.' -ane 'print "$F[0].$F[1] ", join ".", @F[2..$#F]' file 
  3. awk (I'm sure there's a better way)

    awk -F. '{printf "%s.%s ",$1,$2; for(i=3;i<NF;i++){printf "%s.",$(i); }print $NF}' file 

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