I have a number 12345 and I want the result '1,2345'. I tried the following code, but failed:

>>> n = 12345
>>> f"{n:,}"
'12,345'

CodePudding user response：

Regex will work for you:

import re

def format(n):
    return re.sub(r"(\d)(?=(\d{4}) (?!\d))", r"\1,", str(n))

>>> format(12345)
'1,2345'
>>> format(12345678)
'1234,5678'
>>> format(123456789)
'1,2345,6789'

Explanation:

Match:

• (\d) Match a digit...
• (?=(\d{4}) (?!\d)) ...that is followed by one or more groups of exactly 4 digits.

Replace:

• \1, Replace the matched digit with itself and a ,

CodePudding user response：

You can break your number into chunks of 10000's using modulus and integer division, then str.join using ',' delimiters

def commas(n):
    s = []
    while n > 0:
        chunk = n % 10000
        n //= 10000
        s.append(str(chunk))
    return ','.join(reversed(s))

>>> commas(123456789)
'1,2345,6789'
>>> commas(123)
'123'

CodePudding user response：

Sounds like a locale thing^(*). This prints 12,3456,7890 (Try it online!):

import locale

n = 1234567890

locale._override_localeconv["thousands_sep"] = ","
locale._override_localeconv["grouping"] = [4, 0]
print(locale.format_string('%d', n, grouping=True))

That's an I guess hackish way based on this answer. The other answer there talks about using babel, maybe that's a clean way to achieve it.

^(*) Quick googling found this talking about Chinese grouping four digits, and OP's name seems somewhat Chinese, so...

CodePudding user response：

Using babel:

>>> from babel.numbers import format_decimal
>>> format_decimal(1234, format="#,####", locale="en")
'1234'
>>> format_decimal(12345, format="#,####", locale="en")
'1,2345'
>>> format_decimal(1234567890, format="#,####", locale="en")
'12,3456,7890'

This format syntax is specified in UNICODE LOCALE DATA MARKUP LANGUAGE (LDML). Some light bedtime reading there.

Using stdlib only (hackish):

>>> from textwrap import wrap
>>> n = 12345
>>> ",".join(wrap(str(n)[::-1], width=4))[::-1]
'1,2345'