basically helpme.txt has three columns - the first two are decimals and the third is something like 45.5%...

sed is used to discard the % sign as was suggested here earlier...

here is the line of code I'm interested in...

cat helpme.txt | awk '{$1 $2 $3}NR<5{printf("%-20d %-20d %-20.1f\n",$1,$2,$3)}' | sed 's/%//' | sort $3

the output is correct except for the third column - it disappears to 0 values, or rather 0.0

so i am confused. Are there other options I have which don't lose this value? I expected to see 45.5 in the third column of the output. If anyone can give me another hint I would greatly appreciate it as I can't find other alternatives at this point.

-- or is there a way of formatting this output into fixed width columns without it all being the same, ie some fields text and others decimal in the same column?

a sample input and the required output .. would be very useful !!!

The input would be something like

435 1345 45.5%
12345 345676 29%

435            1345            0.0
12345         345675         0.0

Your command works fine for me. I feel the sed isn't necessary and the sort should be based on key - syntax is

sort -k3

$>echo "12 13 45.5%" |awk 'NR<5{printf("%-20d %-20d %-20.1f \n " ,$1,$2,$3)}'|sed 's/%//'
12                   13                   45.5

$>echo "12 13 45.5%" |awk 'NR<5{printf("%-20d %-20d %-20.1f\n",$1,$2,$3)}'
12                   13                   45.5

i cut and pasted your code onto my command line and got the same answer as before...

third column is 0.0

what could be causing this then I'm confused.

Also, when I add a fourth field why does the output double space the lines?

echo "12 13 45.5%" |awk 'NR<5{printf("%-20d %-20d %-20d %-20.1f\n",$1,$1,$2,$3)}'