09-09-2009
Something like this would probably be better done with perl, as you can put the whole matrix in memory.
If you still want to use a shell script, here is one. By the way, I think you meant a symmetric matrix, and not a diagonal matrix.
#!/bin/bash
PATH=/usr/bin:/bin
export PATH
# length of each number in the matrix
rlen=5
# Right-justify the matrix
awk '{ if (ne == "") { ne = NF; } indent = (NR - 1) * (rlen + 1); printf("%" indent "s", ""); print }' rlen="$rlen" > temp.$$
# Fill in the missing parts
cat -n temp.$$ | while read line; do
set -- $line
# Get the row number
n="$1"
# Discard the row number and the 1.000 value
shift 2
# Calculate the start and end positions of the column
# If you're using Bourne shell, you'll have to use expr or similiar.
s=$(( ( $n -1 ) * ( $rlen + 1 ) + 1 ))
e=$(( $s + $rlen - 1 ))
# Get the values of the column for the preceding rows in the matrix
head -$n temp.$$ | cut -c$s-$e | tr '\n' ' '
# Output the rest of the row from the input
echo $*
done
# Clean up
rm -f temp.$$
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