Python: How to convert a number which ll separated by comma as number count increases?

Question

I have a number like: 100 I am showing here as it. But when I am trying to show a number as 1000 then want to display as 1,000.& so on like 1,00,000 likewise.

Below structure

Number Formatted As

10 10

100 100

1000 1,000

10000 10,000

100000 1,00,000

1000000 10,00,000

10000000 1,00,00,000

100000000 10,00,00,000

1000000000 1,00,00,00,000

10000000000 10,00,00,00,000

All the above things I want to do in python.

I thought of using regex but couldn't get an way how to proceed.

Any one having any idea?

So you want the rightmost comma separated digit group to have the width 3 and all others to be two digits wide? — Byte Commander, May 09 '16 at 06:12
@AKS Oops, nice catch. I just assumed, I've never seen numbers separated as provided. — miradulo, May 09 '16 at 06:14
@ByteCommander, looks like the rightmost should be width 3, all else 2. — Alex Huszagh, May 09 '16 at 06:14
@AKS It *is* a duplicate. You need to set the `locale` to `en_IN`. — Bhargav Rao, May 09 '16 at 06:15
@AlexanderHuszagh Oh, yeah. That's what I meant... Confused left and right, duh >.< It's still too early in the morning. — Byte Commander, May 09 '16 at 06:15
@BhargavRao I never said it wasn't. Just that it is not _thousand separators_. `en_IN` it is indeed. — AKS, May 09 '16 at 06:17

Byte Commander · Answer 1 · 2016-05-09T06:55:41.690

Updated: This code now supports both int and float numbers!

You could write a little number-to-string converting function yourself, like this one:

def special_format(n):
    s, *d = str(n).partition(".")
    r = ",".join([s[x-2:x] for x in range(-3, -len(s), -2)][::-1] + [s[-3:]])
    return "".join([r] + d)

It's simple to use:

print(special_format(1))
print(special_format(12))
print(special_format(123))
print(special_format(1234))
print(special_format(12345))
print(special_format(123456))
print(special_format(12345678901234567890))
print(special_format(1.0))
print(special_format(12.34))
print(special_format(1234567890.1234567890))

The above example would result in this output:

1
12
123
1,234
12,345
1,23,456
1,23,45,67,89,01,23,45,67,890
1.0
12.34
1,23,45,67,890.1234567

See this code running on ideone.com

Updated my code snippet to also support float decimal numbers. — Byte Commander, May 09 '16 at 06:56

score 5 · Accepted Answer · answered May 09 '16 at 06:22

5

I recognize this way of separating numbers as the one used in India. So I think you can get what you want using locale:

import locale
locale.setlocale(locale.LC_NUMERIC, 'hi_IN')
locale.format("%d", 10000000000, grouping=True)

The exact locale to use may be different on your system; try locale -a | grep IN to get a list of Indian locales you have installed.

answered May 09 '16 at 06:22

John Zwinck

239,568
38
324
436

I tried with using as: **locale.setlocale(locale.LC_NUMERIC, '')**. Worked fine for me. – danny May 09 '16 at 07:05
@danny: Yeah, that's using the default locale on your machine, which may be fine since you're probably doing this to match your own regional settings. People running your code on systems configured for other countries would need an explicit locale name if they want the to print in "lakh" style. – John Zwinck May 09 '16 at 07:16

Python: How to convert a number which ll separated by comma as number count increases?

2 Answers2