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While-loop with awk

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# 1  
Old 08-28-2007
While-loop with awk
Hi, I have recently posted in another thread started by me Smilie. But in an effort to make my script more beautiful I've been thinking abbout while loops.

I run my script with the command:

sh script 4 numbers.txt

And my script is like this:

Code:
data=`cat $2 | xargs -n $1`

#echo $data

columns=$1
i=1

while columns>0
do
	awk_cmd=`awk '{sum+=$i} END {print "\n" sum/NR}'`
   	echo "$data | $awk_cmd"
	i=`expr $i + 1`
	columns=`expr $columns - 1`
done

The whole point of the script is to take a list ($2) and produce $1 numbers of columns and then print the average of each column. But I get an error when executing the script:

columns: not found

And I don't see the problem. Maybe there are some other issues with the script as well. Please enlighten me.
# 2  
Old 08-28-2007
First, the test condition is wrong. Use this:

Code:
while [ "$columns" -gt 0 ]

Second, if you want to pass the "$i" value to awk try this:

Code:
awk_cmd=`awk -v i=$i '{sum+=i} END {print "\n" sum/NR}'`

# 3  
Old 08-28-2007
Ok I did as you said:

Code:
data=`cat $2 | xargs -n $1`

#echo $data

columns=$1
i=1
echo $columns

while [ "$columns" -gt 0 ]
do
	awk_cmd=`awk -v i=$1 '{sum+=$i} END {print sum/NR}'`
   	echo "$data | $awk_cmd"
	i=`expr $i + 1`
	columns=`expr $columns - 1`
done

But then I get this kind of error:

>> sh reader4 4 testfil.txt
awk: syntax error near line 1
awk: bailing out near line 1
2 2 1 1
1 1 1 1 |
awk: syntax error near line 1
awk: bailing out near line 1
2 2 1 1
1 1 1 1 |
awk: syntax error near line 1
awk: bailing out near line 1
2 2 1 1
1 1 1 1 |
awk: syntax error near line 1
awk: bailing out near line 1
2 2 1 1
1 1 1 1 |

What I want is that is goes through each column on each run in the while loop and prints the corresponding average. But I really appreciate the help, I have been to some forums where you get slammered because, maybe, stupid questions.
# 4  
Old 08-28-2007
You don't need the awk_cmd command :
Code:
data=`cat $2 | xargs -n $1`

#echo $data

columns=$1
i=1
echo $columns

while [ "$columns" -gt 0 ]
do
   	echo "$data" | awk -v i=$1 '{sum+=$i} END {print sum/NR}'
	i=`expr $i + 1`
	columns=`expr $columns - 1`
done

If you want to keep it, you must use the eval command :
Code:
	awk_cmd="awk -v i=$1 '{sum+=$i} END {print sum/NR}'"
	echo "$data" | eval $awk_cmd

Jean-Pierre.
# 5  
Old 08-28-2007
Thanks Jean-Pierre, now I don't get any errors. But the output is wierd.

If I have the list:

1
2
1
2

This would produce:
1 2
1 2

So the output should be 1 and 2 (1+1/2 and 2+2/2 ), like so:
1
2

But instead I get

2
2

Something is wrong with the way I'm going about this problem.

echo "$data" | awk -v i=$1 '{sum+=$i} END {print sum/NR}'

Is there something wrong with using NR here? But putting a number there doesn't seem to help. I tried with 2 instead of NR but (same list as above) but still I get:

2
2

Is there a problem with the division I'm making?

Edit: I just echoed data and the output comes in one line 1 2 1 2 so probably is has something to do with my problem.
Last edited by baghera; 08-28-2007 at 02:50 PM..
# 6  
Old 08-28-2007
I don't quiet understand why you need all the coding when you can do it all in awk:

nawk -f bag.awk myFile

bag.awk:
Code:
{
   for(i=1; i<=NF; i++)
       sum[i] += $i
   nf=NF
   nr=NR
}

END {
    for(i=1; i<=nf; i++)
        print sum[i]/nr
}

Last edited by vgersh99; 08-28-2007 at 03:14 PM..
# 7  
Old 08-28-2007
Quote:
Originally Posted by vgersh99
I don't quot understand why you need all the coding when you can do it all in awk:

nawk -f bag.awk myFile

bag.awk:
Code:
{
   for(i=1; i<=NF; i++)
       sum[i] += $i
   nf=NF
   nr=NR
}

END {
    for(i=1; i<=nf; i++)
        print sum[i]/nr
}

Huh? I tried your script but I only get 1.5

when I have list:

1
2
1
2

The output should be:
1
2

Or on the same row 1 2

As I described above. But there should also be the possibility to tell how many columns there should be.
Last edited by baghera; 08-28-2007 at 03:05 PM..
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