59

In python

 re.sub(r"(?<=.)(?=(?:...)+$)", ",", stroke ) 

To split a number by triplets, e.g.:

 echo 123456789 | python -c 'import sys;import re; print re.sub(r"(?<=.)(?=(?:...)+$)", ",",  sys.stdin.read());'
 123,456,789

How to do the same with bash/awk?

1
  • FWIW, Python has had built-in support for formatting numbers with the comma separator since 2010 (versions 2.7 and 3.1), so you don't need to do the regex trick. Example: print('{:,}'.format(123456789)) #=> 123,456,789
    – Mark Reed
    Nov 14, 2021 at 15:24

14 Answers 14

64

bash's printf supports pretty much everything you can do in the printf C function

type printf           # => printf is a shell builtin
printf "%'d" 123456   # => 123,456

printf from coreutils will do the same

/usr/bin/printf "%'d" 1234567   # => 1,234,567
7
  • 1
    This is now supported in zsh too, updated post here. Dec 13, 2015 at 19:55
  • 1
    I'm on bash 4.1.2 and it doesn't support... :(
    – msb
    Jan 31, 2017 at 19:19
  • 2
    Note printf uses the thousands separator for your current locale, which might be a comma, dot, or nothing at all. You can export LC_NUMERIC="en_US" if you want to force commas.
    – medmunds
    Mar 27, 2017 at 18:31
  • 2
    Get list of supported locale's with locale -a. I had to use en_US.utf8
    – eludom
    May 11, 2018 at 9:48
  • 1
    Watch out for leading zeroes. They will cause printf to treat the number as an octal value (e.g. 01 = 1, 011=9, 0111=73) and return the equivalent decimal. Put another way, this code: num="0123456"; result=$(printf "%'d" "$num"); echo "$result" will return a result of 42,798 and not the expected result of 123,456. May 3, 2021 at 20:09
43

With sed:

$ echo "123456789" | sed 's/\([[:digit:]]\{3\}\)\([[:digit:]]\{3\}\)\([[:digit:]]\{3\}\)/\1,\2,\3/g'
123,456,789

(Note that this only works for exactly 9 digits!)

or this with sed:

$ echo "123456789" | sed ':a;s/\B[0-9]\{3\}\>/,&/;ta'
123,456,789

With printf:

$ LC_NUMERIC=en_US printf "%'.f\n" 123456789
123,456,789
7
  • I'm also trying with awk but it's add comma at the last echo 123456789 | awk '$0=gensub(/(...)/,"\\1,","g")' Feb 6, 2014 at 7:56
  • now I get but it's seems complex echo 123456789 | awk '$0=gensub(/(...)/,"\\1,","g"){sub(",$",""); print}' Feb 6, 2014 at 8:07
  • 1
    That first sed only works if the number is exactly 9 digits. The printf doesn't work on zsh. Thus the second sed answer is probably the best.
    – phemmer
    Feb 6, 2014 at 13:51
  • 1
    @RahulPatil That only works properly if the number of digits is a multiple of 3. Try with "12345678" and you'll see what I mean.
    – phemmer
    Feb 6, 2014 at 13:52
  • 1
    You can do echo 123456789 | awk '{printf ("%'\''d\n", $0)}' (which evidently doesn't always work on Linux!?, but works fine on AIX and Solaris)
    – Johan
    Nov 28, 2014 at 9:36
18

You can use numfmt:

$ numfmt --grouping 123456789
123,456,789

Or:

$ numfmt --g 123456789
123,456,789

Note that numfmt is not a POSIX utility, it is part of GNU coreutils.

4
  • 1
    A neat thing about long options is that they can be abbreviated (as long as the abbreviation is unique).  For example, ls --color can be written ls --col (but not ls --co, because there is also a --context option). It turns out that --grouping is numfmt’s only option that begins with g, so --g is a unique abbreviation. Feb 27, 2020 at 8:37
  • 2
    Just don't use abbreviated forms in scripts, because some day, the one you used might stop working. Sep 13, 2020 at 15:20
  • This does not work for me in coreutils 8.30. (I wonder if LC_ALL=C overrides LC_NUMERIC=en_US.utf8?)
    – RonJohn
    Apr 20, 2021 at 7:50
  • And that was the problem...
    – RonJohn
    Apr 20, 2021 at 7:51
7
cat <<'EOF' |
13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084096
EOF
perl -wpe '1 while s/(\d+)(\d\d\d)/$1,$2/;'

produces:

13,407,807,929,942,597,099,574,024,998,205,846,127,479,365,820,592,393,377,723,561,443,721,764,030,073,546,976,801,874,298,166,903,427,690,031,858,186,486,050,853,753,882,811,946,569,946,433,649,006,084,096

This is accomplished by splitting the string of digits into 2 groups, the right-hand group with 3 digits, the left-hand group with whatever remains, but at least one digit. Then everything is replaced by the 2 groups, separated by a comma. This continues until the substitution fails. The options "wpe" are for error listing, enclose the statement inside a loop with an automatic print, and take the next argument as the perl "program" (see command perldoc perlrun for details).

Best wishes ... cheers, drl

4
  • Thanks to anonymous for the feedback. Even a downvote can be useful, but only if explained -- please comment on what you saw that was wrong. Thanks ... cheers
    – drl
    May 21, 2018 at 15:21
  • I think the downvote here is because you did not explain what the command does. The OP asked for a BASH/AWK alternative so he may not have used PERL before. In any case, best to explain what the command does - especially so for one-liners.
    – AnthonyK
    Jun 3, 2018 at 2:42
  • @AnthonyK -- thanks for probable explanation. I added comments to briefly explain how it works. I think alternative solutions are often useful, but your point about possibly not having used perl is noted ... cheers
    – drl
    Jun 3, 2018 at 10:53
  • I tried the sed and python suggestions on this page. The perl script was the only one that worked for a whole file. The file was filed with text and numbers.
    – Mark
    Aug 24, 2018 at 14:23
5

awk and bash have good built-in solutions, based on printf, as described in the other answers. But first, sed.

For sed, we need to do it "manually". The general rule is that if you have four consecutive digits, followed by a non-digit (or end-of-line) then a comma should be inserted between the first and second digit.

For example,

echo 12345678 | sed -re 's/([0-9])([0-9]{3})($|[^0-9])/\1,\2\3/'

will print

12345,678

We obviously need to then keep repeating the process, in order to keep adding enough commas.

sed -re ' :restart ; s/([0-9])([0-9]{3})($|[^0-9])/\1,\2\3/ ; t restart '

In sed, the t command specifies a label that will be jumped to if the last s/// command was successful. I therefore define a label with :restart, in order that it jumps back.

Here is a bash demo (on ideone) that works with any number of digits:

function thousands {
    sed -re ' :restart ; s/([0-9])([0-9]{3})($|[^0-9])/\1,\2\3/ ; t restart '
}                                                 
echo 12 | thousands
echo 1234 | thousands
echo 123456 | thousands
echo 1234567 | thousands
echo 123456789 | thousands
echo 1234567890 | thousands
1
  • This is like an addon for bash lover, i was strungled with the locale stuff, thankfully found this gem. very practical, just need to paste this one line of code somewhere inside the shell, and use it whenever needed. Btw, it's hard to understand what the code does, the only thing i know is i can replace the ”,” with ”.” Mar 16 at 16:17
3

With some awk implementations:

echo "123456789" | awk '{ printf("%'"'"'d\n",$1); }'  

123,456,789  

"%'"'"'d\n" is: "%(single quote)(double quote)(single quote)(double quote)(single quote)d\n"

That will use the configured thousand separator for your locale (typically , in English locales, space in French, . in Spanish/German...). Same as returned by locale thousands_sep

1
  • 1
    a maybe cleaner way is : awk '{printf("%\047d\n",$1); }', and awk will translate the octal 047 into a single quot before interpreting the printf string, avoiding you this weird string of quote and doublequotes. You can force a specific locale (if it is installed...) : LC_NUMERIC=en_US awk .... (or fr_FR if you want spaces, and if it is installed) Sep 5, 2020 at 15:45
2

A common use case for me is to modify the output of a command pipeline so that decimal numbers are printed with thousand separators. Rather than writing a function or script, I prefer to use a technique that I can customise on the fly for any output from a Unix pipeline.

I have found printf (provided by Awk) to be the most flexible and the memorable way to to accomplish this. The apostrophe/single quote character is specified by POSIX as a modifier to format decimal numbers and has the advantage that it’s locale-aware so it’s not restricted to using comma characters.

When running Awk commands from a Unix shell, there can be difficulties entering a singe-quote character inside a string delimited by single-quotes (to avoid shell expansion of positional variables, e.g., $1). In this case, I find the most readable and reliable way to enter the single-quote character is to enter it as an octal escape sequence (beginning with \0).

Example:

printf "first 1000\nsecond 10000000\n" |
  awk '{printf "%9s: %11\047d\n", $1, $2}'
  first:       1,000
 second:  10,000,000

Simulated output of a pipeline showing which directories are using the most disk space:

printf "7654321 /home/export\n110384 /home/incoming\n" |
  awk '{printf "%22s: %9\047d\n", $2, $1}'
  /home/export: 7,654,321
/home/incoming:   110,384

Other solutions are listed in How to escape a single quote inside awk.

Note: as warned against in Print a Single Quote, it’s recommended to avoid the use of hexadecimal escape sequences as they do not work reliably across different systems.

2
  • 1
    Of all the awk-based answers listed on here, this one is most certainly the most graceful (IMHO). One doesn't need to hack in a quote with other quotes like in other solutions. Apr 3, 2019 at 6:57
  • Thanks @TSJNachos117 The hardest part is remembering that the octal encoding for the apostrophe character is \047. Apr 24, 2019 at 12:24
1
$ echo 1232323 | awk '{printf(fmt,$1)}' fmt="%'6.3f\n"
12,32,323.000
1

If you are looking at BIG numbers I was unable to make the above solutions work. For example, lets get a really big number:

$ echo 2^512 |bc -l|tr -d -c [0-9] 13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084096

Note I need the tr to remove backslash newline output from bc. This number is too big to treat as a float or fixed bit number in awk, and I don't even want to build a regexp large enough to account for all the digits in sed. Rather, I can reverse it and put commas between groups of three digits, then unreverse it:

echo 2^512 |bc -l|tr -d -c [0-9] |rev |sed -e 's/\([0-9][0-9][0-9]\)/\1,/g' |rev 13,407,807,929,942,597,099,574,024,998,205,846,127,479,365,820,592,393,377,723,561,443,721,764,030,073,546,976,801,874,298,166,903,427,690,031,858,186,486,050,853,753,882,811,946,569,946,433,649,006,084,096

1
  • 2
    Good answer. However, I've never encountered a problem using large numbers with Awk. I tried your example on a number of Red Hat and Debian-based distributions but in all cases, Awk had no problem with the large number. I thought some more about it and it occurred to me that all the systems I had experimented on were 64-bit (even a very old VM running unsupported RHEL 5). It wasn’t until I tested an old lap-top running a 32-bit OS that I was able to replicate your issue: awk: run time error: improper conversion(number 1) in printf("%'d. Jan 11, 2019 at 21:49
1
a="13407807929942597099574024998205846127479365820592393377723561443721764030073546976801874298166903427690031858186486050853753882811946569946433649006084096"

echo "$a" | rev | sed "s#[[:digit:]]\{3\}#&,#g" | rev

13,407,807,929,942,597,099,574,024,998,205,846,127,479,365,820,592,393,377,723,561,443,721,764,030,073,546,976,801,874,298,166,903,427,690,031,858,186,486,050,853,753,882,811,946,569,946,433,649,006,084,096
2
  • That adds a spurious leading comma if the number of digits in the number is a multiple of 3. Jun 3, 2018 at 8:09
  • @StéphaneChazelas: You could take the output of that last rev command, and pipe it to sed 's/^,//g'. Apr 3, 2019 at 6:51
1

A bash/awk (as requested) solution that works regardless of the length of the number and uses , regardless of the locale's thousands_sep setting, and wherever the numbers are in the input and avoids adding the thousand separator after in 1.12345:

echo not number 123456789012345678901234567890 1234.56789 |
  awk '{while (match($0, /(^|[^.0123456789])[0123456789]{4,}/))
        $0 = substr($0, 1, RSTART+RLENGTH-4) "," substr($0, RSTART+RLENGTH-3)
        print}'

Gives:

not number 123,456,789,012,345,678,901,234,567,890 1,234.56789

With awk implementations like mawk that don't support the interval regex operators, change the regexp to /(^|[^.0123456789])[0123456789][0123456789][0123456789][0123456789]+/

1
  • This is great for using with any shell output, thanks.
    – kilves76
    Dec 24, 2019 at 6:54
0

The following uses space as thousands separator, which is the practice at my place. Modifying it for using comma should be easy.

echo "1000066955"|sed -rn "s/([[:digit:]])([[:digit:]]{3})$/\1 \2/;T end;:loop s/([[:digit:]])([[:digit:]]{3})[[:space:]]/\1 \2 /;t loop;:end p;"
0

I also wanted to have the part after the decimal separator correctly separated/spaced, therefore I wrote this sed-script which uses some shell variables to adjust to regional and personal preferences. It also takes into account different conventions for the number of digits grouped together:

#DECIMALSEP='\.' # usa                                                                                                               
DECIMALSEP=','   # europe
    
#THOUSSEP=','  # usa
#THOUSSEP='\.' # europe
#THOUSSEP='_'  # underscore
#THOUSSEP=' '  # space
THOUSSEP=' '   # thinspace
    
# group before decimal separator
#GROUPBEFDS=4   # china
GROUPBEFDS=3    # europe and usa
    
# group after decimal separator
#GROUPAFTDS=5   # used by many publications 
GROUPAFTDS=3
    
    
function digitgrouping {
# FIXME: This is a workaround: BEGINNING has to be marked (and after                                                                
# alteration removed) for the first number to be spaced correctly (1234
# should be 1 234, and that only works if something is in front of that
# number).
sed -e 's%^%BEGINNING&%' \
  -e '
  s%\([0-9'"$DECIMALSEP"']\+\)'"$THOUSSEP"'%\1__HIDETHOUSSEP__%g
  :restartA ; s%\([0-9]\)\([0-9]\{'"$GROUPBEFDS"'\}\)\(['"$DECIMALSEP$THOUSSEP"']\)%\1'"$THOUSSEP"'\2\3% ; t restartA
  :restartB ; s%\('"$DECIMALSEP"'\([0-9]\{'"$GROUPAFTDS"'\}\'"$THOUSSEP"'\)*\)\([0-9]\{'"$GROUPAFTDS"'\}\)\([0-9]\)%\1\3'"$THOUSSEP"'\4% ; t restartB
  :restartC ; s%\([^'"$DECIMALSEP"'][0-9]\+\)\([0-9]\{'"$GROUPBEFDS"'\}\)\($\|[^0-9]\)%\1'"$THOUSSEP"'\2\3% ; t restartC
  s%__HIDETHOUSSEP__%\'"$THOUSSEP"'%g' \
  -e 's%^BEGINNING%%'

}
6
  • This will fail in the UK. Why not just use the Locale? There are no separators after the decimal point. Jun 23, 2021 at 18:49
  • 1
    @JeremyBoden: By Europe I mean European Union.
    – erik
    Jun 24, 2021 at 10:45
  • 1
    The thousand separator is different in Spain (.), France (space) and Ireland (,) which are all part of the EU. Jun 24, 2021 at 11:49
  • 1
    See also locale thousands_sep and locale -k LC_NUMERIC (on GNU systems at least) Jun 24, 2021 at 11:50
  • Note that it assumes GNU sed. POSIXly and with several sed implementations, you can't have other commands after branching ones (:, t, b...). \+ is also a GNU extension. Jun 24, 2021 at 11:54
0
= Number grouping formatting using Perl RegEx =

[
|*| Source: https://unix.stackexchange.com/a/656655
|*| Last update: CE 2021-08-18 06:44 UTC ]


Number grouping formatting (e.g. turning "1000000" into "1,000,000"; approximation of `numfmt --grouping`) using Perl RegEx:
(Unix Shell)
[
    PERLIO=':raw:utf8' exec '/usr/bin/perl' -p \
    -e 'BEGIN { $^H |= 0x02800000; $^H{reflags_charset} = 4; $/ = undef(); }' \
    -e '

    sub f {
    $x1 = $1;
    $x2 = $2;
#
# [
    if ( length( $x1 ) > 3 ) {
    pos( $x1 ) = length( $x1 ) % 3;
    $x1 =~ s/\G.{3}/ ( pos( $x1 ) != 0 ? "," : "" ).${&}; /gse;
    };
# ]
#
# Would work but inefficient:
# [
#   $x1 =~ s/(?<=\d)(?=(\d+))/ ( length( $1 ) % 3 != 0 ? "" : "," ); /ge;
# ]
# ,
# [
#   $x1 =~ s/(?<=\d)(?=(?:\d{3})+(?!\d))/,/g;
# ]
#
    "${x1}${x2}";
    };

    s/(?<![\w#&)*,.\/:;=-\@\[-\]`{-}])([0-9]+)(\.[0-9]+)?(?![\w#\$&(*\-\/<=\@\[-\]`{-}]|\.[^\W0-9])/ f(); /geu;

    ' \
    "$@";
]
[ Explanation Needed ]


Test case:
(Console Log (Unix) )
[
> \
    { nf <<\EOF
0.000000
10.000000
100.000000
1000.000000
10000.000000
100000.000000
1000000.000000
10000000.000000
100000000.000000
1000000000.000000
10000000000.000000
100000000000.000000
1000000000000.000000
EOF
    } | nf; # Verified idempotence.

0.000000
10.000000
100.000000
1,000.000000
10,000.000000
100,000.000000
1,000,000.000000
10,000,000.000000
100,000,000.000000
1,000,000,000.000000
10,000,000,000.000000
100,000,000,000.000000
1,000,000,000,000.000000
]
[ Alternatively: Try the full text of this message. ]




See also:
|*| "perlrun" - how to execute the Perl interpreter # "-i''[extension]''": https://perldoc.perl.org/perlrun#-i%5Bextension%5D
2
  • Welcome to the site, and thank you for your contribution. Please use the proper formatting tools for your post, as currently it is very difficult to separate explanations from actual code.
    – AdminBee
    Jul 2, 2021 at 7:34
  • @AdminBee: pastebin.com/wDmNQJpD Jul 2, 2021 at 13:06

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