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(linux) man ls:
If that is not absolutely essential, tryfor i in *
do echo item: $i
if [ -s line${i}_Ux.xy ]
then awk '{a+=$2; n++;} END {print a/n}' $i
else...

The for-in-loop should not be used. The OP wants to preserve the order of the list, buti in a is not guaranteed to preserve order.

awk '{sum+=$2; a[NR]=$1} END{for (i=1; i<=NR; i++) {print...

That said looking at your script now, do you mind explaining what $( ls -v ) is supposed to give and what OS we are dealing with ( and the shell used perhaps?) because -v is not implemented on AIX or...

you are missing a fi, thats all, just before the last line...


for i in $( ls -v )
do
echo item: $i
if [ ! -s line$i_Ux.xy ]; then
echo -e "0" >> list
else
cat...

awk '{sum+=$2;a[++i]=$1} END{for (i in a) {print a[i]"/"sum}}' infile

$ awk 'NR==FNR{a+=$2;next}{print $1"/"a}' file file
1/5
2/5
3/5
4/5

formula in example does not show average calculation, but try:
awk 'NR==1 {a=$1} {b+=$2; c=$1} END {print (b/NR) * (c - a)}' input

This awk code might help:
awk '
{
if ( $1 in L )
{
L[$1] = L[$1] > $2 ? $2 : L[$1]
H[$1] = H[$1] < $2 ?...

Here is another awk approach:
awk '
{
if ( $1 in L )
{
L[$1] = L[$1] > $2 ? $2 : L[$1]
H[$1] = H[$1] < $2...

Try:awk ' {if (!L[$1] && $2 < L[$1]) L[$1]=$2
if (!H[$1] && $2 > H[$1]) H[$1]=$2}
END {for (x in L) print x, L[x], H[x]}
' file
1 -0.1 0.1
2 -0.2...

Just create two variables and load them with the first column value (min and max so far). On lines 2-$ check to see if you have a new min, and if not, do you have a new max (can't be both!)?

Try:awk '{s[$1]+=$2;n[$1]++}END{for (i in s) print i,s[i]/n[i]}' list.txt