获取Python十进制的精确十进制字符串表示形式? [英] Get precise decimal string representation of Python Decimal?
问题描述
如果我有Python Decimal
,如何可靠地获取数字的精确十进制字符串(即,不是科学计数法)表示形式而不尾随零? / p>
例如,如果我有:
>>> ; d =小数('1e-14')
我想要:
>> get_decimal_string(d)
'0.00000000000001'
但是:
-
十进制
类没有任何to_decimal_string
方法,甚至任何to_radix_string(radix)
(cf: https://docs.python.org/3/library/decimal.html#decimal.Context.to_eng_string ) - The
%f
格式化程序默认默认舍入到小数点后6位-'%f'%(d,)==> '0.000000'
-或要求精确的小数位数。 -
{:f} .format(...)
格式化程序出现起作用-'{:f}'。format(d)
-但是我不愿意相信这一点,因为这实际上与文档,其中显示
==> '0.00000000000001''f'
…显示为固定数字,点精度。默认精度为6 -
十进制。__repr __
和十进制。__str __
有时返回科学计数法:repr(d)==> Decimal('1E-14')
因此,有什么方法可以获取小数Python 十进制
中的字符串?还是我需要使用 Decimal.as_tuple()
自己滚动?
简短答案:
>> d
小数('1E-14')
>>> '{:f}'。format(d)
'0.00000000000001'
长答案:
如布兰登罗兹指出的 PEP 3101 (PEP的字符串格式)状态:
格式说明符的语法是开放式的,因为类可以使
覆盖标准格式说明符。在这种情况下,
str.format()方法仅将
第一个冒号和匹配括号之间的所有字符传递给相关的基础
格式化方法。
因此, Decimal .__ format __
方法是python的字符串格式将用来生成<$ 十进制
值的c $ c> str 表示形式。基本上十进制
会将格式覆盖为智能格式,但默认为格式字符串设置的任何值(即 {:. 4f}
会将小数位截断为4位。
这就是为什么您可以信任它(摘录自 decimal.py:Decimal .__ format __
):
def __format __(self,specifier,context = None, _localeconv = None):
#
#...实施被截断
#
#找出小数点的位置
leftdigits = self._exp + len(self._int)
如果spec ['type']在' eE':
如果不是self且精度不为None:
dotplace = 1-精度
else:
dotplace = 1
elif spec ['type'] in 'fF%':
dotplace =左位数
elif spec ['type'] in'gG':
if self._exp< = 0且左位数> -6:
dotplace =左位数
else:
dotplace = 1
#查找小数点前后的数字,如果dotplace<,则得到指数
; 0:
intpart ='0'
fracpart ='0'*(-dotplace)+ self._int
elif dotplace> len(self._int):
intpart = self._int +'0'*(dotplace-len(self._int))
fracpart =''
else:
intpart = self._int [:dotplace]或'0'
fracpart = self._int [dotplace:]
exp = leftdigits-dotplace
#用十进制特定的东西完成;将格式的其余
#交给_format_number函数
返回_format_number(self._sign,intpart,fracpart,exp,spec)
长话短说, Decimal .__ format __
方法将计算必要的填充,以表示小数点前后的数字基于 Decimal._exp
提供的幂运算(在您的示例中为14个有效数字)。
>> d._exp
-14
If I've got a Python Decimal
, how can I reliably get the precise decimal string (ie, not scientific notation) representation of the number without trailing zeros?
For example, if I have:
>>> d = Decimal('1e-14')
I would like:
>>> get_decimal_string(d)
'0.00000000000001'
However:
- The
Decimal
class doesn't have anyto_decimal_string
method, or even anyto_radix_string(radix)
(cf: https://docs.python.org/3/library/decimal.html#decimal.Context.to_eng_string) - The
%f
formatter either defaults to rounding to 6 decimal places -'%f' %(d, ) ==> '0.000000'
- or requires a precise number of decimal places. - The
{:f}.format(...)
formatter appears to work -'{:f}'.format(d) ==> '0.00000000000001'
- however I'm reluctant to trust that, as this actually runs counter to the documentation, which says "'f'
… Displays the number as a fixed-point number. The default precision is 6" Decimal.__repr__
andDecimal.__str__
sometimes return scientific notation:repr(d) ==> "Decimal('1E-14')"
So, is there any way to get a decimal string from a Python Decimal
? Or do I need to roll my own using Decimal.as_tuple()
?
Short answer:
>>> d
Decimal('1E-14')
>>> '{:f}'.format(d)
'0.00000000000001'
Long answer:
As Brandon Rhodes pointed out PEP 3101 (which is the string format PEP) states:
The syntax for format specifiers is open-ended, since a class can override the standard format specifiers. In such cases, the str.format() method merely passes all of the characters between the first colon and the matching brace to the relevant underlying formatting method.
And thus, the Decimal.__format__
method is what python's string format will utilize to generate the str
representation of the Decimal
value. Basically Decimal
overrides the formatting to be "smart" but will default to whatever values the format string sets (ie {:.4f}
will truncate the decimal to 4 places).
Here's why you can trust it (snippet from decimal.py:Decimal.__format__
):
def __format__(self, specifier, context=None, _localeconv=None):
#
# ...implementation snipped.
#
# figure out placement of the decimal point
leftdigits = self._exp + len(self._int)
if spec['type'] in 'eE':
if not self and precision is not None:
dotplace = 1 - precision
else:
dotplace = 1
elif spec['type'] in 'fF%':
dotplace = leftdigits
elif spec['type'] in 'gG':
if self._exp <= 0 and leftdigits > -6:
dotplace = leftdigits
else:
dotplace = 1
# find digits before and after decimal point, and get exponent
if dotplace < 0:
intpart = '0'
fracpart = '0'*(-dotplace) + self._int
elif dotplace > len(self._int):
intpart = self._int + '0'*(dotplace-len(self._int))
fracpart = ''
else:
intpart = self._int[:dotplace] or '0'
fracpart = self._int[dotplace:]
exp = leftdigits-dotplace
# done with the decimal-specific stuff; hand over the rest
# of the formatting to the _format_number function
return _format_number(self._sign, intpart, fracpart, exp, spec)
Long story short, the Decimal.__format__
method will calculate the necessary padding to represent the number before and after the decimal based upon exponentiation provided from Decimal._exp
(in your example, 14 significant digits).
>>> d._exp
-14
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