Floating-point numbers are represented in computer hardware as base 2 (binary) fractions. For example, the decimal fraction:
0.125
has value 1/10 + 2/100 + 5/1000, and in the same way the binary fraction:
0.001
has value 0/2 + 0/4 + 1/8. These two fractions have identical values, the only real difference being that the first is written in base 10 fractional notation, and the second in base 2.
浮点数在计算机硬件中以二进制分数表示。 例如，十进制分数：
0.125
包含1/10 + 2/100 + 5/1000的值，同样二进制分数:
0.001
包含0/2 + 0/4 + 1/8的值。这两种分数具有同样的值，实际唯一的区别就是第一种以十进制表示书写，而第二种则以二进制表示。
Unfortunately, most decimal fractions cannot be represented exactly as binary fractions. A consequence is that, in general, the decimal floating-point numbers you enter are only approximated by the binary floating-point numbers actually stored in the machine.
不幸的，大多十进制分数无法用二进制分数精确的表示。 通常，一种后果就是你输入的十进制浮点数存储在机器中实际上只能用近似的二进制浮点数表示。
The problem is easier to understand at first in base 10. Consider the fraction 1/3. You can approximate that as a base 10 fraction:
0.3
or, better,
0.33
or, better,
0.333
and so on. No matter how many digits you’re willing to write down, the result will never be exactly 1/3, but will be an increasingly better approximation of 1/3.
首先，这个问题在十进制中比较容易理解。 思考1/3这个分数，你可以将它约等于十进制的分数。
0.3
精确一点：
0.3
或者更精确一点：
0.3
等等。不论你愿意写多少数字，结果永远不会精确的等于1/3，只能是更近似于1/3。
In the same way, no matter how many base 2 digits you’re willing to use, the decimal value 0.1 cannot be represented exactly as a base 2 fraction. In base 2, 1/10 is the infinitely repeating fraction.
同样的，不论你愿意用多少二进制数字，十进制值0.1都无法用二进制分数精确的表示。 在二进制中，1/10是无限循环的分数。
0.0001100110011001100110011001100110011001100110011...
Stop at any finite number of bits, and you get an approximation. On most machines today, floats are approximated using a binary fraction with the numerator using the first 53 bits following the most significant bit and with the denominator as a power of two. In the case of 1/10, the binary fraction is 3602879701896397 / 2 ** 55 which is close to but not exactly equal to the true value of 1/10.
你可以通过有限的位数获取一个近似值。 在当今的大多数机器上，浮点数都是使用一个带有53位有效位的二进制分数作为分子和以分母为指数的2的方作为分母表示的近似值。 以1/10来说，二进制分数为``3602879701896397 / 2 ** 55``，只能近似表示而无法精确的等于1/10的值。
Many users are not aware of the approximation because of the way values are displayed. Python only prints a decimal approximation to the true decimal value of the binary approximation stored by the machine. On most machines, if Python were to print the true decimal value of the binary approximation stored for 0.1, it would have to display。
大多数用户都因为值的显示方法而未曾意识到这个近似值。 Python只把在机器中存储为二进制近似值的真实的十进制值输出为一个十进制的近似值。 在大多数机器中，如果Python能用真实的十进制值输出一个二进制近似值存储的0.1，它就会以此显示。
>>> 0.1
0.1000000000000000055511151231257827021181583404541015625
That is more digits than most people find useful, so Python keeps the number of digits manageable by displaying a rounded value instead.
这比大多数用户发现的有用值更数字化，所以为易于管理Python通过舍入值来代替显示数字化的数字。
>>> 1 / 10
0.1
Just remember, even though the printed result looks like the exact value of 1/10, the actual stored value is the nearest representable binary fraction.
记住，即使输出结果看上去是1/10的精确值，实际上存储的值是最接近的二进制分数。
Interestingly, there are many different decimal numbers that share the same nearest approximate binary fraction. For example, the numbers 0.1 and 0.10000000000000001 and 0.1000000000000000055511151231257827021181583404541015625 are all approximated by 3602879701896397 / 2 ** 55. Since all of these decimal values share the same approximation, any one of them could be displayed while still preserving the invariant eval(repr(x)) == x.
有趣的是，有很多不同的十进制数字共享同一个最接近的二进制分数。 例如，数字``0.1``和``0.10000000000000001``和``0.1000000000000000055511151231257827021181583404541015625``都近似的表示为``3602879701896397 / 2 ** 55``。 因为所有这些十进制值共享同一个近似表示，它们中的任何一个都可以显示成保持不变的``eval(repr(x)) == x``。
Historically, the Python prompt and built-in repr() function would chose the one with 17 significant digits, 0.10000000000000001, Starting with Python 3.1, Python (on most systems) is now able to choose the shortest of these and simply display 0.1.
以往，Python提示符和内建:func:`repr`函数可能选择带有17位有意义的数字``0.10000000000000001``。 从Python 3.1开始，Python（在大多数系统上）现在可以从中选择最短的表示并简单的显示``0.1``。
Note that this is in the very nature of binary floating-point: this is not a bug in Python, and it is not a bug in your code either. You’ll see the same kind of thing in all languages that support your hardware’s floating-point arithmetic (although some languages may not display the difference by default, or in all output modes).
注意：这在二进制浮点数中是很自然的：这不是Python的一个bug，也不是你代码中的一个bug。 你可以在所有支持你的硬件的浮点计算的语言中遇见这个相同的问题（尽管有些语言默认可能不会*显示*它们的区别，甚至在所有的输出中都一样）。
Python’s built-in str() function produces only 12 significant digits, and you may wish to use that instead. It’s unusual for eval(str(x)) to reproduce x, but the output may be more pleasant to look at.
Python的内置函数:func:`str`仅显示12位有效数字，或许你希望选择使用它。 使用``eval(str(x))``复制*x*是不寻常的，但它的输出看上去可能更舒服。
>>> str(math.pi)
'3.14159265359'
>>> repr(math.pi)
'3.141592653589793'
>>> format(math.pi, '.2f')
'3.14'
It’s important to realize that this is, in a real sense, an illusion: you’re simply rounding the display of the true machine value.
事实上，一定要认识到这只是一个幻觉：你只是将机器值简单的舍入*显示*了。
One illusion may beget another. For example, since 0.1 is not exactly 1/10, summing three values of 0.1 may not yield exactly 0.3, either.
一种幻觉可能引发另一种幻觉。 列如，因为0.1不是精确的1/10，将3个0.1相加可能不会产生精确的0.3。
>>> .1 + .1 + .1 == .3
False
Also, since the 0.1 cannot get any closer to the exact value of 1/10 and 0.3 cannot get any closer to the exact value of 3/10, then pre-rounding with round() function cannot help.
同时，因为0.1不可能无限的接近1/10精确的值，0.3不可能无限的接近3/10精确的值，所以使用:func:`round`函数预先舍入也不会相等。
>>> round(.1, 1) + round(.1, 1) + round(.1, 1) == round(.3, 1)
False
Though the numbers cannot be made closer to their intended exact values, the round() function can be useful for post-rounding so that results have inexact values that are comparable to one another.
尽管这些数字不能接近它们想要表示的精确值，:func:`round`函数可以使用稍后舍入，所以结果具有不精密值，这便可以与其它的比较。
>>> round(.1 + .1 + .1, 1) == round(.3, 1)
True
Binary floating-point arithmetic holds many surprises like this. The problem with “0.1” is explained in precise detail below, in the “Representation Error” section. See The Perils of Floating Point for a more complete account of other common surprises.
二进制浮点运算还有许多类似的惊奇。 “0.1”的问题在下面“Representation Error”小节中有更准确的解释。 更多完整的惊奇请参考：`The Perils of Floating Point <http://www.lahey.com/float.htm>`_
As that says near the end, “there are no easy answers.” Still, don’t be unduly wary of floating-point! The errors in Python float operations are inherited from the floating-point hardware, and on most machines are on the order of no more than 1 part in 2**53 per operation. That’s more than adequate for most tasks, but you do need to keep in mind that it’s not decimal arithmetic, and that every float operation can suffer a new rounding error.
就像在结尾说的那样：“没有简单的答案。” 然而，无需过度的为浮点数忧虑！ Python浮点操作中的错误都是继承自浮点硬件，并且在大多数机器上每个操作不超过1/2**53。 那足以胜任大多数任务，但你必须意识到它不是十进制运算，并且每个浮点数操作都可能遭受一个新的舍入错误。
While pathological cases do exist, for most casual use of floating-point arithmetic you’ll see the result you expect in the end if you simply round the display of your final results to the number of decimal digits you expect. str() usually suffices, and for finer control see the str.format() method’s format specifiers in Format String Syntax.
虽然确实存在这个问题，如果你简单将最终的显示结果舍入为期望的十进制数字，对大多数偶尔使用的浮点运算最后你都会看到期望的结果。 通常:func:`str`函数即可做到，如果想要更好的控制请参考:ref:`formatstrings`文档中:meth:`str.format`方法的格式化分类符。
For use cases which require exact decimal representation, try using the decimal module which implements decimal arithmetic suitable for accounting applications and high-precision applications.
对于那些需要精确十进制表示的情况，可以尝试使用:mod:`decimal`模块，它为会计和高精度要求类应用程序实现了十进制运算。
Another form of exact arithmetic is supported by the fractions module which implements arithmetic based on rational numbers (so the numbers like 1/3 can be represented exactly).
另一个支持精确运算的是:mod:`fractions`模块，它实现了基于有理数的运算（因此像1/3这样的数字就可以被精确的表示）。
If you are a heavy user of floating point operations you should take a look at the Numerical Python package and many other packages for mathematical and statistical operations supplied by the SciPy project. See <http://scipy.org>.
如果你需要大量使用浮点操作，那么你应该参考*Numerical*的Python包，或者有*SciPy*项目提供的其他许多支持数学和统计学操作的包。 参考<http://scipy.org>。
Python provides tools that may help on those rare occasions when you really do want to know the exact value of a float. The float.as_integer_ratio() method expresses the value of a float as a fraction.
当你确切的想知道一个浮点数的精确值时，Python为此提供了有用的工具。 :meth:`float.as_integer_ratio`函数将一个浮点数的值表示为一个分数。
>>> x = 3.14159
>>> x.as_integer_ratio()
(3537115888337719L, 1125899906842624L)
Since the ratio is exact, it can be used to losslessly recreate the original value.
因为分数的比是精确的，它可以用来无损的创建原始值。
>>> x == 3537115888337719 / 1125899906842624
True
The float.hex() method expresses a float in hexadecimal (base 16), again giving the exact value stored by your computer.
| meth: | `float.hex`方法将一个浮点数表示为十六进制（以16为基数），并且给出存储在你计算机中的精确值。 |
|---|
>>> x.hex()
'0x1.921f9f01b866ep+1'
This precise hexadecimal representation can be used to reconstruct the float value exactly.
这个精确的十六进制表示可以用来精确的重建浮点数值。
>>> x == float.fromhex('0x1.921f9f01b866ep+1')
True
Since the representation is exact, it is useful for reliably porting values across different versions of Python (platform independence) and exchanging data with other languages that support the same format (such as Java and C99).
因为这个表示是精确的，它可以在不同的Python版本（平台独立）间可靠的传递值，并且与其他支持相同格式的语言（像Java和C99）交换数据。
Another helpful tool is the math.fsum() function which helps mitigate loss-of-precision during summation. It tracks “lost digits” as values are added onto a running total. That can make a difference in overall accuracy so that the errors do not accumulate to the point where they affect the final total.
另外一个有用的工具是:func:`math.fsum`函数，它用来在加法中减少精度损失。 它跟踪被加到一个运行总数上的作为值的“丢失数字”。 那会影响总的精度，因此错误不会堆积到影响最终总数的点上。 :
>>> sum([0.1] * 10) == 1.0
False
>>> math.fsum([0.1] * 10) == 1.0
True
This section explains the “0.1” example in detail, and shows how you can perform an exact analysis of cases like this yourself. Basic familiarity with binary floating-point representation is assumed.
Representation error refers to the fact that some (most, actually) decimal fractions cannot be represented exactly as binary (base 2) fractions. This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many others) often won’t display the exact decimal number you expect.
Why is that? 1/10 is not exactly representable as a binary fraction. Almost all machines today (November 2000) use IEEE-754 floating point arithmetic, and almost all platforms map Python floats to IEEE-754 “double precision”. 754 doubles contain 53 bits of precision, so on input the computer strives to convert 0.1 to the closest fraction it can of the form J/2**N where J is an integer containing exactly 53 bits. Rewriting
1 / 10 ~= J / (2**N)
as
J ~= 2**N / 10
and recalling that J has exactly 53 bits (is >= 2**52 but < 2**53), the best value for N is 56:
>>> 2**52
4503599627370496
>>> 2**53
9007199254740992
>>> 2**56/10
7205759403792794.0
That is, 56 is the only value for N that leaves J with exactly 53 bits. The best possible value for J is then that quotient rounded:
>>> q, r = divmod(2**56, 10)
>>> r
6
Since the remainder is more than half of 10, the best approximation is obtained by rounding up:
>>> q+1
7205759403792794
Therefore the best possible approximation to 1/10 in 754 double precision is that over 2**56, or
7205759403792794 / 72057594037927936
Dividing both the numerator and denominator by two reduces the fraction to:
3602879701896397 / 36028797018963968
Note that since we rounded up, this is actually a little bit larger than 1/10; if we had not rounded up, the quotient would have been a little bit smaller than 1/10. But in no case can it be exactly 1/10!
So the computer never “sees” 1/10: what it sees is the exact fraction given above, the best 754 double approximation it can get:
>>> 0.1 * 2 ** 55
3602879701896397.0
If we multiply that fraction by 10**60, we can see the value of out to 60 decimal digits:
>>> 3602879701896397 * 10 ** 60 // 2 ** 55
1000000000000000055511151231257827021181583404541015625
meaning that the exact number stored in the computer is approximately equal to the decimal value 0.100000000000000005551115123125. Rounding that to 17 significant digits gives the 0.10000000000000001 that Python displays (well, will display on any 754-conforming platform that does best-possible input and output conversions in its C library — yours may not!).
The fractions and decimal modules make these calculations easy:
>>> from decimal import Decimal
>>> from fractions import Fraction
>>> print(Fraction.from_float(0.1))
3602879701896397/36028797018963968
>>> print(Decimal.from_float(0.1))
0.1000000000000000055511151231257827021181583404541015625