0

I have a process which outputs too many status-lines with carriage return (\r). I can filter all those status lines by piping them through

sed '/\r/d'

I would instead like to filter all of these lines except, e.g. every 3. Is this possible with standard Unix-Tools (awk?) or do I need a script for that? Lines without CR should be left untouched.

Given Output:

$ (printf '%s\n' {1..10};   printf  '%s\r\n' {1..10}; printf '%s\n' {1..10};)  | cat -v
1
2
3
4
5
6
7
8
9
10
1^M
2^M
3^M
4^M
5^M
6^M
7^M
8^M
9^M
10^M
1
2
3
4
5
6
7
8
9
10

Wanted output (or any other pattern):

1
2
3
4
5
6
7
8
9
10
1^M
4^M
7^M
10^M
1
2
3
4
5
6
7
8
9
10
David Weber
  • 11

3 Answers3

1
$ awk '!(/\r$/ && ((++c)%3 != 1))' file | cat -v
1
2
3
4
5
6
7
8
9
10
1^M
4^M
7^M
10^M
1
2
3
4
5
6
7
8
9
10

Original answer:

Sounds like all you need is this, using any awk:

awk -v RS='\r' '{ORS=(NR%10000 ? "" : RS)} 1'

e.g. using this as input:

$ printf '%s\r\n' {1..10} | cat -v
1^M
2^M
3^M
4^M
5^M
6^M
7^M
8^M
9^M
10^M

Removing all but every 3rd \r:

$ printf '%s\r\n' {1..10} | awk -v RS='\r' '{ORS=(NR%3 ? "" : RS)} 1' | cat -v
1
2
3^M
4
5
6^M
7
8
9^M
10
Ed Morton
  • 31,617
0

Using GNU sed, we use the hold space for counting.

sed -E '
  /\r$/{
    G;/\n$/P
    s/.*\n/./
    /.{3}/z;x;d
  }
' file

Using awk, we Use the variable c as a circular counter that gets reset whenver it reaches 3.

awk '
!/\r$/ || !c++
c==3{c=0}
' file

Assuming the carriage returns (\r) , whenever they occur, occur at the end of a line feed (\n) delimited record.

guest_7
  • 5,728
  • 1
  • 7
  • 13
-1

Here's a fringe way to do it in awk :

{m,g}awk '((+$_ % 3) % NF)~(!_<NF)' FS='\r$'  # yes that's a 
                                              # tilde ~ not a minus -

1
2
3
4
5
6
7
8
9
10
1^M
4^M
7^M
10^M
1
2
3
4
5
6
7
8
9
10

Other ways of saying the same thing

mawk 'NF-!_== (+$+_   %    3    ) % NF' FS='\r$'
gawk 'NF-!_== ( $(_++)%(_+_+_--)) % NF' FS='\r$'
RARE Kpop Manifesto
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