python - 如何打印带有数千个分隔符的浮点数？

Question

如何格式化十进制数字以便32757121.33显示为32.757.121,33？

Answer 1

使用locale.format()：

>>> 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数值的显示（使用 等时locale.format()）locale.str()，而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

Answer 2

我找到了另一个解决方案：

'{:,.2f}'.format(num).replace(".","%").replace(",",".").replace("%",",")

Answer 3

如果由于某种原因不能或不想使用locale，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()采用标准 Python 浮点数的字符串表示形式，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'

解释：

\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

Answer 4

>>> import locale

>>> locale.setlocale(locale.LC_ALL, 'en_GB.UTF-8')
'en_GB.UTF-8'
>>> print(locale.format_string('%.2f', 12742126.15, True))
12,742,126.15

此示例代码适用于 docker 容器中的 GB。

FROM python:3.8.2-slim-buster

RUN apt-get update && apt-get install -y locales && \
    sed -i -e 's/# en_GB.UTF-8 UTF-8/en_GB.UTF-8 UTF-8/' /etc/locale.gen && \
    dpkg-reconfigure --frontend=noninteractive locales

ENV LANG en_GB.UTF-8
ENV LC_ALL en_GB.UTF-8

可以通过在终端上运行以下命令来找到语言环境 (Linux Dirstro)

locale -a

然后出现完整的语言环境列表：

en_AG.utf8
en_AU.utf8
...
en_GB.utf8
...