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Transposing rows and columns (pivoting) using shell scripting

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# 8  
Old 04-24-2012
I should begin with reading scripting tutorials/books and visit these forums.

Consider the follow code:
Code:
awk -F, 'NR==1{n=split($0,a,FS);print $1;next}{for(i=2;i<=n;i++)print $1,a[i],$i}' file

Should an explanation of the code above make any sense to you? Maybe I'm wrong but I doubt it.....
Last edited by Franklin52; 04-24-2012 at 10:35 AM..
# 9  
Old 04-25-2012
I would be glad if you explain me your code. Please!
# 10  
Old 04-25-2012
Quote:
Originally Posted by ksatish89
I would be glad if you explain me your code. Please!
If you never used awk an explanation shouldn't make any sense like I mentionend on my last post.
But if you persist:
Code:
awk -F, 'NR==1{n=split($0,a,FS);print $1;next}{for(i=2;i<=n;i++)print $1,a[i],$i}' file

Explanation:

awk -F, use a comma as field separator
'NR==1{n=split($0,a,FS);print $1;next} if NR==1 use the split function to get the values of the 1st fields in array a, print the 1st field and get the next record.
{for(i=2;i<=n;i++)print $1,a[i],$i}' loop from field 2 and print the 1st field the value of the element a[i] and field $i



Have a read of an awk tutorial for the used commands:

Awk - A Tutorial and Introduction - by Bruce Barnett
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