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Awk: for i in 1,2,n-1,n


 
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# 1  
Old 09-16-2013
Awk: for i in 1,2,n-1,n

hello,

a fast question:

how can i loop over four numbers in awk like in bash:

n ist a number >3

Code:
n=4
for i in 1 2 $((n-1)) $n
do .. done

i tried:

awk: for (i in 1 2 n-1 n)
with all possible brackets (){} and punctuations , ;
but this doesn't work...
# 2  
Old 09-16-2013
Put the values in an array and to iterate in order use a for loop with an index variable.

Regards,
Alister
This User Gave Thanks to alister For This Post:
# 3  
Old 09-16-2013
try a loop like:
Code:
echo "" | awk '{x=4 ; while (i++ <x) print i}'

# 4  
Old 09-16-2013
You could try somehting like this:
Code:
awk 'BEGIN{n=4; s="1,2,"n-1; split(s,a,","); for (i in a) print a[i]}'

This User Gave Thanks to Subbeh For This Post:
# 5  
Old 09-16-2013
EDIT : The solution below does not fit the question, sorry about that :X

Is this OK?
Code:
~/unix.com$ awk 'BEGIN{n=4; for(i=1; i<=n; i++)print i}'
1
2
3
4
~/unix.com$ awk 'BEGIN{n=4; while(i++<n)print i}'
1
2
3
4


Last edited by tukuyomi; 09-16-2013 at 01:27 PM..
# 6  
Old 09-16-2013
Maybe not quite what you need, but at least pointing in the desired direction:
Code:
awk '{for (i=n-2; i<=n+1; i++) print i%n+1} ' n=15 file
14
15
1
2

A little sort might help...
This User Gave Thanks to RudiC For This Post:
# 7  
Old 09-16-2013
problem solved

@alister: works, but not very elegant

@tukuyomi, rdrtx1: n can be >>> 4
i need the first two (1,2) and the last two (n-1,n)...

@rudiC:
elegant, modulo! the order is no problem

@Subbeh: could work, I will try...

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