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'