thousands separator

ynixon · #1 (**permalink**) 02-12-2008

Hi,
Trying to represent a number with thousands separator in AWK:

Code:

echo 1 12 123 1234 12345 123456 1234567 | awk --re-interval '{print gensub(/([[:digit:]])([[:digit:]]{3})/,"\\1,\\2","g")}' 

  1 12 123 1,234 1,2345 1,23456 1,234567

any idea what is wrong here ?

radoulov · #2 (**permalink**) 02-12-2008

I would use this (as it seams you have GNU Awk):

Code:

% cat sep.awk
{ printf "%'d ", $1 } END { print "" }
% print 1 12 123 1234 12345 123456 1234567 |gawk -f sep.awk RS=" "
1 12 123 1,234 12,345 123,456 1,234,567

Your locale must support such characters:

Code:

% print 1 12 123 1234 12345 123456 1234567 |LC_ALL=C gawk -f sep.awk RS=" "
1 12 123 1234 12345 123456 1234567
% print 1 12 123 1234 12345 123456 1234567 |LC_ALL=en_US.UTF-8 gawk -f sep.awk RS=" "
1 12 123 1,234 12,345 123,456 1,234,567

ynixon · #3 (**permalink**) 02-12-2008

I get the following output:

Code:

%'d %'d %'d %'d %'d %'d %'d

I am using redhat release 4

radoulov · #4 (**permalink**) 02-12-2008

I suppose it's version/environment specific.

A User’s Guide for GNU Awk
Edition 3
June, 2004

Quote:

’ A single quote or apostrohe character is a POSIX extension to ISO C. It indicates
that the integer part of a floating point value, or the entire part of an
integer decimal value, should have a thousands-separator character in it. This
only works in locales that support such characters. For example:
$ cat thousands.awk # Show source program
a BEGIN { printf "%’d\n", 1234567 }
$ LC_ALL=C gawk -f thousands.awk # Run it in "C" locale
a 1234567
$ LC_ALL=en_US.UTF-8 gawk -f thousands.awk # Run in US English
UTF locale
a 1,234,567

ynixon · #5 (**permalink**) 02-12-2008

maybe it works for the manual
still it doesn't work for me

radoulov · #6 (**permalink**) 02-12-2008

Quote:

Originally Posted by ynixon

maybe it works for the manual
still it doesn't work for me

Not only for the manual, it works fine on my Ubuntu 7.10

gllo · #7 (**permalink**) 04-12-2008

Quote:

Originally Posted by ynixon

maybe it works for the manual
still it doesn't work for me

The

Code:

printf "%'d "

solution did not work for me either. I have GNU AWK 3.1.5. The man doesn't mention apostrophe among printf format options, and I don't have thousands.awk file.

This solution

Code:

echo 1 12 123 1234 12345 123456 1234567 | awk --re-interval '{print gensub(/([[:digit:]])([[:digit:]]{3})/,"\\1,\\2","g")}'

will not work, because only the #,### pattern gets repeated. It becomes more clear when you add a few longer numbers to the list.

I created the following solution:

Code:

#!/bin/sh 
nums=`echo -e " 1\n 12\n 123\n 1234\n 12345\n 123456\n 1234567\n 12345678\n 123456789\n 1234567890\n"`
echo "$nums" | awk --re-interval '{ 
        if (length($1) > 3) 
        {
                a = int(length($1)%3)
                
                if (a == 0)
                {
                        p1 = gensub(/([[:digit:]]{3})/, "\\1,", "g")
                        printf "%-20d %s \n", $1, gensub(/,$/, "\\1", "g", p1)
                }

                if (a == 1)
                {
                        q1 = gensub(/\<([[:digit:]])/, "\\1,", "g")
                        q2 = gensub(/([[:digit:]]{3})/, "\\1,", "g", q1)
                        printf "%-20d %s \n", $1, gensub(/,$/, "\\1", "g", q2)
                }
                
                if (a == 2)
                {
                        r1 = gensub(/\<([[:digit:]]{2})/, "\\1,", "g")
                        r2 = gensub(/([[:digit:]]{3})/, "\\1,", "g", r1)
                        printf "%-20d %s \n", $1, gensub(/,$/, "\\1", "g", r2)
                }
        }
}'

Note! This will not work with non-integers (i don't need it for my script), but it can be extended with some effort!

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