How to restrict the decimal places in awk without rounding off

Question

I want to print 0.74 for 0.748.

If I use

awk '{printf "%0.2f\n",$1}' filename

for 0.748 it prints 0.75. But I want 0.74 and remove remaining part. For 77.348 it should be 77.34.

Any ideas?

There is no

gawk -M

option to use ROUNDMODE="Z" ??

Even I couldn't find ceiling and floor functions as well.

Have you considered printing it as a string e.g. "%.4s\n"? — steeldriver, Jul 13 '15 at 11:14
@steeldriver Consider there are two values 4.567 and 77.657 for example. Now using %.1s in printf prints 4 and 7 respectively. But I need it as 4 and 77. It is not taking care of dot here — Ramana Reddy, Jul 13 '15 at 11:20
If you just want the integer part, you can use the %d conversion specifier. On 15.04 with GNU awk 4.1.1, it is possible to use MPFR to set the ROUNDMODE explicitly - see Setting the Rounding Mode but AFAIK that's not available in other versions/varieties — steeldriver, Jul 13 '15 at 11:31

score 1 · Accepted Answer · answered Jul 15 '15 at 16:17

Using awk

with print

awk -F. '{print $1"."substr($2,1,2)}' filename

or

with printf

awk -F. '{printf "%0.2f\n",$1"."substr($2,1,2)}' filename

Example

$ awk -F. '{print $1"."substr($2,1,2)}' foo
0.74
77.65

$ awk -F. '{printf "%0.2f\n",$1"."substr($2,1,2)}' foo
0.74
77.65

$ cat foo
0.748
77.657

1 Answers1