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Shell Programming and Scripting

Print String Every Specific Line

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# 1  
Old 02-04-2013
Print String Every Specific Line
Dear All,

I have input file like this,

001
059
079
996
758
079
069
059
079
...
...

Desired output:

AA 001
BB 059
CC 079
DD 996
AA 758
BB 079
CC 069
DD 059
AA 079
...
...

print different string (AA,BB,CC,DD) every multiples of four.....

Thanks Advance,


Attila
# 2  
Old 02-04-2013
Try something like:
Code:
awk 'BEGIN {
        p[0] = "DD"
        p[1] = "AA"
        p[2] = "BB"
        p[3] = "CC"
}
{       printf("%s %s\n", p[NR % 4], $0)}' input

As always, if you are using a Solaris/SunOS system, use /usr/xpg4/bin/awk or nawk instead of awk.
# 3  
Old 02-05-2013
Or sth like this..

Code:
 
awk 'BEGIN{split("DD AA BB CC",P)}
{print P[NR%4+1],$0}' file

# 4  
Old 02-05-2013
Quote:
Originally Posted by pamu
Or sth like this..

Code:
 
awk 'BEGIN{split("DD AA BB CC",P)}
{print P[NR%4+1],$0}' file

Not quite. In awk, + has higher precedence than % (so you're doing mod 5 rather than (mod 4) + 1). If you change the subscript in the 2nd line to (NR%4)+1, you'd get what you wanted.

-------------------------------------
Ouch... I apologize. I was looking at unary + not binary +. Pamu's code is correct as written.

Sorry,
Don
# 5  
Old 02-05-2013
you can use this also

Code:
  sed -e "1~4s/^/AA /g" -e "2~4s/^/BB /g" -e "3~4s/^/CC /g" -e "4~4s/^/DD /g" $file

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