awk - set numbers [ 1 ... n] from the 6 line


 
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# 1  
Old 01-14-2016
awk - set numbers [ 1 ... n] from the 6 line

Hi,

i have a file, where measurement-data is stored in the first column. The file has also a header of 5 lines. I want to set counting up numbers [1, 2, 3, ...] in front of any particular measurement-value; should start at the 6. line with starting number 1.
i try to solve it with ...
Code:
awk 'NR > 6 { print FNR "\t" $0 }' file.dat

..but it will cut the header. Can someone give me a hint, how to give out the complete file?

Thanks in advance,
IMPe
# 2  
Old 01-14-2016
Hello IMPe,

Request you to please always do show sample input and expected sample output too. If I understood correctly your requirement, could you please try following and let me know if that helps.
Code:
###If you want to leave empty line.
awk 'NR > 6 && NF{ print ++i "\t" $0 }'  Input_file
 
###If you want to count each line then following may help.
awk 'NR > 6 { print ++i "\t" $0 }'  Input_file

Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
# 3  
Old 01-14-2016
Some sample always helps! Try (untested)
Code:
awk 'NR>5 {sub(/^/, NR-5)} 1' file

This User Gave Thanks to RudiC For This Post:
# 4  
Old 01-14-2016
Quote:
Originally Posted by RavinderSingh13
Hello IMPe,

Request you to please always do show sample input and expected sample output too. If I understood correctly your requirement, could you please try following and let me know if that helps.
Code:
###If you want to leave empty line.
awk 'NR > 6 && NF{ print ++i "\t" $0 }'  Input_file
 
###If you want to count each line then following may help.
awk 'NR > 6 { print ++i "\t" $0 }'  Input_file

Thanks,
R. Singh
Hi RavinderSingh,

thank you for the fast reply. Unfortunately your one-liner (both) work the same way and cut the top of the header like my awk-example, too. The given awk-one-liner, from the following threadpost, of RudiC work propper.

Thank you!
IMPe

---------- Post updated at 06:53 AM ---------- Previous update was at 06:51 AM ----------

Quote:
Originally Posted by RudiC
Some sample always helps! Try (untested)
Code:
awk 'NR>5 {sub(/^/, NR-5)} 1' file

Hi RudiC,

thanks a lot; Your one-liner is working propper and do exactly what i want.

Thank you!
IMPe
# 5  
Old 01-14-2016
Quote:
Originally Posted by IMPe
Hi RavinderSingh,

thank you for the fast reply. Unfortunately your one-liner (both) work the same way and cut the top of the header like my awk-example, too. The given awk-one-liner, from the following threadpost, of RudiC work propper.

Thank you!
IMPe

---------- Post updated at 06:53 AM ---------- Previous update was at 06:51 AM ----------



Hi RudiC,

thanks a lot; Your one-liner is working propper and do exactly what i want.

Thank you!
IMPe
Hello IMPe,

Please try following it should print the headers too, not tested though.
Code:
###If you want to leave empty line.
awk 'NR==1{print;next} NR > 6 && NF{ print ++i "\t" $0 }'  Input_file
 
###If you want to count each line then following may help.
awk 'NR==1{print;next} NR > 6 { print ++i "\t" $0 }'  Input_file

Thanks,
R. Singh
# 6  
Old 01-14-2016
The following prints in cat -n style
Code:
awk 'NR>h {printf "%6s ",NR-h} {print}' h=5 file.dat

# 7  
Old 01-14-2016
Quote:
Originally Posted by MadeInGermany
The following prints in cat -n style
And why don't use cat -n (or less -N, or nl) Smilie

Code:
head -n5 file; (tail -n+6 file | cat -n)

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