如何格式化十进制数,以便32757121.33
显示为32.757.121,33
?
答案 0 :(得分:16)
>>> import locale
>>> locale.setlocale(locale.LC_ALL, 'German')
'German_Germany.1252'
>>> print(locale.format('%.2f', 32757121.33, True))
32.757.121,33
您可以将区域设置更改限制为显示数值(使用locale.format()
,locale.str()
等时)并保持其他区域设置不受影响:
>>> locale.setlocale(locale.LC_NUMERIC, 'English')
'English_United States.1252'
>>> print(locale.format('%.2f', 32757121.33, True))
32,757,121.33
>>> locale.setlocale(locale.LC_NUMERIC, 'German')
'German_Germany.1252'
>>> print(locale.format('%.2f', 32757121.33, True))
32.757.121,33
答案 1 :(得分:10)
我找到了另一个解决方案:
'{:,.2f}'.format(num).replace(".","%").replace(",",".").replace("%",",")
答案 2 :(得分:5)
如果由于某种原因你不能或不想使用locale
,你也可以使用正则表达式:
import re
def sep(s, thou=",", dec="."):
integer, decimal = s.split(".")
integer = re.sub(r"\B(?=(?:\d{3})+$)", thou, integer)
return integer + dec + decimal
sep()
获取标准Python float的字符串表示形式,并使用自定义的千位和小数分隔符返回它。
>>> s = "%.2f" % 32757121.33
>>> sep(s)
'32,757,121.33'
>>> sep(s, thou=".", dec=",")
'32.757.121,33'
<强>解释强>
\B # Assert that we're not at the start of the number
(?= # Match at a position where it's possible to match...
(?: # the following regex:
\d{3} # 3 digits
)+ # repeated at least once
$ # until the end of the string
) # (thereby ensuring a number of digits divisible by 3