Format numbers using bash?

KClaisse · 09-24-2008, 06:11 AM

I am trying to find a way to convert large numbers into comma separated numbers using bash. I.e. 12345 -> 12,345 I tried using the printf solution: printf "%'d\n" 12345678 but it doesn't seem to work in Cygwin (probably because locale isn't supported yet). Is there a way to do this within bash?

ChrisAbela · 09-24-2008, 06:50 AM

num=12345
echo "${num%???},${num#??}"

ghostdog74 · 09-24-2008, 07:22 AM

Quote:

Originally Posted by ChrisAbela

num=12345
echo "${num%???},${num#??}"

this is not enough. num may not be 12345 everytime.

theYinYeti · 09-24-2008, 07:25 AM

What exactly are you trying to achieve? How should 12345678 be rendered?

Yves.

ghostdog74 · 09-24-2008, 07:26 AM

Quote:

Originally Posted by KClaisse

I am trying to find a way to convert large numbers into comma separated numbers using bash. I.e. 12345 -> 12,345 I tried using the printf solution: printf "%'d\n" 12345678 but it doesn't seem to work in Cygwin (probably because locale isn't supported yet). Is there a way to do this within bash?

if you have Python and can use it, here's an alternative

Code:

#!/usr/bin/env python
import locale
locale.setlocale(locale.LC_ALL,('en','ascii')) 
print locale.format('%.2f', 5554435.555, 3)

output

Code:

# ./test.py
5,554,435.55

theYinYeti · 09-24-2008, 07:35 AM

Oh I just understood! In most of the world (or so I've read), and at least where I live, the comma stands for the decimal separator. I suppose “your” comma is a thousands separator, isn't it?

Yves.

chrism01 · 09-24-2008, 07:41 AM

In the English speaking world your 2nd sentence is the case

PS re OP: That does work in bash:

printf "%'d\n" 12345678
12,345,678

KClaisse · 09-25-2008, 02:05 AM

Quote:

Originally Posted by chrism01

In the English speaking world your 2nd sentence is the case

PS re OP: That does work in bash:

printf "%'d\n" 12345678
12,345,678

Unfortunately that relies on locale which isn't available in Cygwin, although it works fine on a normal box. I'm just looking to format the output of any number, no decimals among then, to include comma's for presentation purposes. There are thousands of numbers so I cannot do them by hand. Here's an example:

1 = 1
10 = 10
100 = 100
1000 = 1,000
10000 = 10,000
100000 = 100,000
1000000 = 1,000,000

etc.

archtoad6 · 09-25-2008, 08:29 AM

I think this will do the trick for you, unless your Cygwin doesn't have sed or, possibly, GNU sed:

Code:

$ echo " 1234567890 " | sed -r ':L;s=\b([0-9]+)([0-9]{3})\b=\1,\2=g;t L'
 1,234,567,890

Thank you for the interesting Q -- I got me to look at & use sed labels & loops for the 1st time.

Anticipating your Q's:
http://www.gnu.org/software/sed/manual/sed.html
":L": http://www.gnu.org/software/sed/manu...mming-Commands
"-r": http://www.gnu.org/software/sed/manu...02dextended-30
"\b": http://www.gnu.org/software/sed/manual/sed.html#Escapes
"=": http://www.gnu.org/software/sed/manu...s_0022-Command (para 1)

(Cheshire Cat imitation)

KClaisse · 09-25-2008, 11:22 AM

Bravo sir, that works perfectly. It will take me a while to wrap my head around though

. Thank you very much.

archtoad6 · 09-27-2008, 05:42 AM

You're welcome.

As to "wrapping your head around it", I figured it might be unfamiliar which is why I gave all the links. I'll be happy to answer any specific Q's.

turtlegeek · 02-23-2011, 08:36 PM

So I spent a few minutes to find the solution below. Since my Mac only allows Basic Regular Expression (BRE), I had to change a few things and add a few backslashes to escape the control characters:

Code:

echo "1234567890" | \
sed -e ': L
s/\([0-9]\{1,19\}\)\([0-9]\{3\}\)/\1,\2/
t L'

Note that the backslash on the first line must immediately precede a newline
and the limit of 19 for the first RE can be changed to the maximum number of
digits you may encounter. I haven't tried this on ubuntu yet but it should
work for any system that accepts BRE's (and has sed).

To see how the command works, remove the branch (t) command. You'll see that
the first iteration puts the last (rightmost) comma in. Each successive
loop matches the left portion of the number without commas. The last loop
exits because there's nothing left to match.

Reuti · 02-24-2011, 04:24 AM

Quote:

Originally Posted by turtlegeek

So I spent a few minutes to find the solution below. Since my Mac only allows Basic Regular Expression (BRE)...

In such cases I sometimes download the GNU version and compile it on the Mac. Several commands have slight variations in the GNU version.

archtoad6 · 02-24-2011, 08:14 AM

turtlegeek,

Welcome to LQ -- good 1st post.

Thanks for the translation to "BRE". If nothing else, you show why it is important to some of us to seek out a ver. of sed that has the "-r" option available. (Thanks, Reuti.)
Technically, it's not that your Mac doesn't have "extended regular expressions", it's the ver. of sed which it supplies.
Please look into & use [code] blocks. BTW, if you like, you can edit your post to add them.
Good use of the backslash & separate lines. It's clearer that way.
I re-ran & re-tested my orig. code & I find that the spaces in the input string & the '\b'-s used to find & clear them were unnecessary.
In GNU sed the "-e" is unnecessary.
Unless your ver. of sed doesn't allow it, your sed expression would be clearer if you used "s=...=...=", instead of "s/.../.../". -- Given the number of backslashes necessary to convert to BRE, it becomes a "kung fu regex"
I assume that your sed doesn't allow "[0-9]+", so you used [0-9]{1,19}. BTW, it could have been "[0-9]{4,19}".
My "g" option is unnecessary.
Good point about the backslash on the first line.
Great explanation of how it works.

Thanks for bringing this back to my attention, I had almost as much fun reviewing & rethinking it as I did originally.

Reuti,

Thanks for your suggestion.

My tested updated ver.:

Code:

echo "1234567890"  | \
sed -r '
  :L
  s=([0-9]+)([0-9]{3})=\1,\2=
  t L'

archtoad6 · 02-24-2011, 08:16 AM

acc-dupe trying to edit.