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Indexing each repeating pattern of rows in a column using awk/sed

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# 1  
Old 09-07-2014
Indexing each repeating pattern of rows in a column using awk/sed
Hello All,

I have data like this in a column.
Code:
0
1
2
3
0
3
4
5
6
0
1
2
3
etc.

where 0 identifies the start of a pattern in my data.
So I need the output like below using either awk/sed.
Code:
0  1
1  1
2  1
3  1
0  2
3  2
4  2
5  2
6  2
0  3
1  3
2  3
3  3

Basically indexing every repeating pattern and add additional column with that index number.

Thanks in advance
Sidda
# 2  
Old 09-07-2014
As I am sure you already know, sed doesn't do arithmetic...
Try:
Code:
awk '
$1 == 0{ pat++ }
{ printf("%s  %d\n", $0, pat) }
' data

If you want to try this on a Solaris/SunOS system, change awk to /usr/xpg4/bin/awk, /usr/xpg6/bin/awk, or nawk.
This User Gave Thanks to Don Cragun For This Post:
ks_reddy
# 3  
Old 09-07-2014
Or
Code:
awk '{printf "%s  %d\n", $0, p+=$1==0}' file
0  1
1  1
2  1
3  1
0  2
3  2
4  2
5  2
6  2
0  3
1  3
2  3
3  3

or even
Code:
awk '{print $0, p+=$1==0}' file

This User Gave Thanks to RudiC For This Post:
ks_reddy
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