mjhjmtu

I have a bash script that prints the following results:

120,900
1160,001
80,730
600,165
6,310
1111,203

I would like to left pad that so the result will be:

 120,900
1160,001
 80,730
 600,165
 6,310
1111,203

I already use that line to keep just 3 numbers after the comma in the second column

awk 't" "%.3fn", $2 ' MyFile.txt;

How can I accomplish this?

Use printf rather than echo:

$ cat 442817.sh
#!/bin/bash
numbers=(120,900 1160,001 80,730 600,165 6,310 1111,203)
for n in "$numbers[@]"; do
 printf "%10sn" "$n"
done
$ ./442817.sh
 120,900
 1160,001
 80,730
 600,165
 6,310
 1111,203

Awk solution:

awk 'FNR == NR len = length; if (len > max_len) max_len = len; next 
 printf "%" max_len "sn", $0 ' file.txt file.txt

The output:

 120,900
1160,001
 80,730
 600,165
 6,310
1111,203

Or via GNU coreutils:

printf "%$(sort -nr file.txt | wc -L)sn" $(cat file.txt)

$ column -t -s, file | sed 's/^([0-9]*)( *)/21,/'
 120,900
 1160,001
 80,730
 600,165
 6,310
 1111,203

The column output will be

120 900
1160 001
80 730
600 165
6 310
1111 203

and the sed command just moves the spaces between the two columns to the start of the line and inserts the comma in their place.

If the initial extra two spaces are not wanted, use

$ column -t -s, file | sed -e 's/^([0-9]*)( *)/21,/' -e 's/^ //'
 120,900
1160,001
 80,730
 600,165
 6,310
1111,203