I need to printf a number out, but with given width and rounded (with awk!)


I have this and somehow I need to connect the %d but everything I do, ends up with too much parametres for awk (because I have more columns).


You can try this:

$ awk 'BEGIN{printf "%3.0f\n", 3.6}'

Our format option has two parts:

  • 3: meaning output will be padded to 3 characters.
  • .0f: meaning output will have no precision, meaning rounded up.

From man awk, you can see more details:

width   The field should be padded to this width. The field is normally padded
        with spaces. If the 0  flag  has  been  used, it is padded with zeroes.

.prec   A number that specifies the precision to use when printing.  For the %e,
        %E, %f and %F, formats, this specifies the number of digits you want
        printed to the right of the decimal point. For the %g, and %G formats,
        it specifies the maximum number of significant  digits. For the %d, %o,
        %i, %u, %x, and %X formats, it specifies the minimum number of digits to
        print. For %s, it specifies the maximum number of characters from the
        string that should be printed.

Using the %f format specifier, your (floating point) number will get automatically rounded as you specify. For example, to round a value to whole numbers use

$ awk 'BEGIN { printf("%.0f\n", 1.49); }'
$ awk 'BEGIN { printf("%.0f\n", 1.5); }'

If you want further trailing digits, just change precision.

  • is /dev/null necessary? – Avinash Raj May 20 '14 at 7:44
  • If your only statements are in the BEGIN block, it's not. I tested with the expression in the normal body first, thus mea culpa. Thanks, @Gnouc. – Andreas Wiese May 20 '14 at 7:54

Awk uses sprintf underneath and it does unbiased rounding, so depending on your platform if you want it to ALWAYS round up you may need to use something like this:

awk "BEGIN { x+=(5/2); printf('%.0f', (x == int(x)) ? x : int(x)+1) }"

Not realizing this can lead to subtle but nasty bugs.

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