如何对python小数进行区域设置格式并保持其精度? [英] How can I locale-format a python Decimal and preserve its precision?

问题描述

如何使用语言环境格式化十进制?

How can I format a Decimal using a locale?

为了举例说明，我试图定义一个函数 f ，使其在美国英语语言环境中:

To describe by examples, I'm trying to define a function f such that in the US English locale:

• f(十进制('5000.00'))=='5,000.00'

f(十进制('1234567.000000'))=='1,234,567.000000'

一些不起作用的东西:

• f = str 不使用语言环境； f(Decimal('5000.00'))=='5000.00'

f = lambda d:locale.format('％f'，d)不保留小数精度； f(Decimal('5000.00'))=='5000.000000'

f = lambda d:locale.format('％.2f'，d)使用固定精度，这不是我想要的； f(Decimal('1234567.000000'))=='1234567.00'

f = lambda d: locale.format('%.2f', d) uses a fixed precision, which isn't what I'm after; f(Decimal('1234567.000000')) == '1234567.00'

推荐答案

通读 Decimal 模块的源代码， Decimal .__ format __ 提供完整的

Reading through the source for the decimal module, Decimal.__format__ provides full PEP 3101 support, and all you have to do is select the correct presentation type. In this case, you want the :n type. According PEP 3101 spec, this :n has the following properties:

'n'-数字.这与'g'相同，但它使用的是当前的语言环境设置以插入适当的数字分隔符.

'n' - Number. This is the same as 'g', except that it uses the current locale setting to insert the appropriate number separator characters.

这比其他答案更简单，并且避免了我原来的答案中的浮点精度问题(保留在下面):

This is simpler than other answers, and avoids the float precision issue in my original answer (preserved below):

>>> import locale
>>> from decimal import Decimal
>>> 
>>> def f(d):
...     return '{0:n}'.format(d)
... 
>>> 
>>> locale.setlocale(locale.LC_ALL, 'en_us')
'en_us'
>>> print f(Decimal('5000.00'))
5,000.00
>>> print f(Decimal('1234567.000000'))
1,234,567.000000
>>> print f(Decimal('123456700000000.123'))
123,456,700,000,000.123
>>> locale.setlocale(locale.LC_ALL, 'no_no')
'no_no'
>>> print f(Decimal('5000.00'))
5.000,00
>>> print f(Decimal('1234567.000000'))
1.234.567,000000
>>> print f(Decimal('123456700000000.123'))
123.456.700.000.000,123

原始错误答案

您可以告诉格式字符串使用与小数本身一样多的精度，并使用语言环境格式化程序:

Original, wrong answer

You can just tell the format string to use as much precision as is included in the decimal itself and use the locale formatter:

def locale_format(d):
    return locale.format('%%0.%df' % (-d.as_tuple().exponent), d, grouping=True)

请注意，如果您拥有一个与实数相对应的小数，则可以使用，但是如果小数是 NaN 或 + Inf 或其他形式，则无法正常工作像那样.如果您的输入中有这些可能性，则需要使用format方法进行解释.

Note that works if you've got a decimal which corresponds to a real number, but doesn't work correctly if the decimal is NaN or +Inf or something like that. If those are possibilities in your input, you'd need to account for them in the format method.

>>> locale.setlocale(locale.LC_ALL, 'en_US')
'en_US'
>>> locale_format(Decimal('1234567.000000'))
'1,234,567.000000'
>>> locale_format(Decimal('5000.00'))
'5,000.00'
>>> locale.setlocale(locale.LC_ALL, 'no_no')
'no_no'
>>> locale_format(Decimal('1234567.000000'))
'1.234.567,000000'
>>> locale_format(Decimal('5000.00'))
'5.000,00'
>>> locale_format(Decimal('NaN'))
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in locale_format
TypeError: bad operand type for unary -: 'str'

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