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Sum of three columns - in 4N columns file

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# 8  
Old 01-07-2010
Thank you for all the reply.
I got from Radoulov the idea of using awk with -3 option although I didn't know what -3 meant. So any other solution is welcome.
I tried the solution

Code:
awk '/^[0-9]/ && $1==0 && (N=NF/4) { for (i=1;i<=N;i++) $i=sprintf("%.4E", $(N+i)+$(2*N+i)+$(3*N+i))} -3' OFS='\t' data

but I get as the output the input itself.

I tried also the first solution of alister which he showed to be working but I still get as the output the input itself.

Maybe the =0 is not detected with floating point?

---------- Post updated at 04:33 AM ---------- Previous update was at 04:16 AM ----------

Sorry there was a mistake in the input file (1 extra tab before the first column). now it's fixed.
thank you again,
sarah
# 9  
Old 01-07-2010
Quote:
Originally Posted by alister
Hello ahmad.diab:

The number of columns is 4*N. The problem statement says that they are divided into four groups. The only columns that will be modified to contain the sum of their counterparts are the first N. Your code is going beyond the first N columns and modifying ones it shouldn't.

Take care,
alister
Thanks alister for the clarification but as you can see f_o_555 post was not
very clear to me, and as per my code was modified and used by you it means
that I was on the correct track and then misunderstand the post.

Thanks again man

SmilieSmilieSmilieSmilie
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