Shell Programming and Scripting

Count up an index (e.g. CNT) for every new $2 occurring, and then, in the END section, don't use (f in K) but (i=1; i<=CNT; i++).
Or, use NR==1 to print the header.

Rudi:

When you say "count up an index" I am not sure what is meant. I am only familar with simple counter scripts. Do you mean the following?

Code:

CNT=0
{count[$2]++; keyword[$2] = $1}
END {
for (i=1;i<=CNT;i++)

or would it be CNT=$2? Pardon my ignorance.

No, sth like (untested)

Code:

{if !($2 in count) CNT++; count[$2]++; keyword[$2] = $1} 
END { for (i=1;i<=CNT;i++)

so, for every $2 not yet found in the array, CNT will be incremented and can then be used in the END section.

This was so simple, it irks me! I took a look at the output, and as I checked carefully, the files were being processed in reverse order!! All I did then was run the sort command with the script and it was all solved. Can some awk guru tell me why awk processed files in reverse order?

Here is what I did:

Code:

for f in F*; do gawk -f script_0113b $f |sort ;done

This is the file:

Code:

{count[$2]++; keyword[$2] = $1}
END {
for (k in count)
{if (count[k] == 2) keyword[k] = "rearranging"
if (count[k] == 3) keyword[k] = "deleting"
if (count[k] > 3)  print k ": Volumes has more than 3 repeats" > "/dev/stderr"
else print keyword[k] " " k
}
}

If I run the code without sort, the files are jumbled up and are actually in reverse order so they aren't correct for what I need to do.


It appears also that the following command does not work the same as sort -r. Does anyone know why?

Code:

awk '{a[i++]=$0} END {for (j=i-1; j>=0;) print a[j--] }'

The files are not processed in reversed order, but for k in count produces the array elements in an undetermined order..