14. 浮点算术：争议和限制¶
浮点数在计算机硬件中表示为以 2 为基数（二进制）的小数。举例而言，十进制的小数
0.125
等于 1/10 + 2/100 + 5/1000 ，同理，二进制的小数
0.001
等于0/2 + 0/4 + 1/8。这两个小数具有相同的值，唯一真正的区别是第一个是以 10 为基数的小数表示法，第二个则是 2 为基数。
不幸的是，大多数的十进制小数都不能精确地表示为二进制小数。这导致在大多数情况下，你输入的十进制浮点数都只能近似地以二进制浮点数形式储存在计算机中。
用十进制来理解这个问题显得更加容易一些。考虑分数 1/3 。我们可以得到它在十进制下的一个近似值
0.3
或者，更近似的，:
0.33
或者，更近似的，:
0.333
以此类推。结果是无论你写下多少的数字，它都永远不会等于 1/3 ，只是更加更加地接近 1/3 。
同样的道理，无论你使用多少位以 2 为基数的数码，十进制的 0.1 都无法精确地表示为一个以 2 为基数的小数。 在以 2 为基数的情况下， 1/10 是一个无限循环小数
0.0001100110011001100110011001100110011001100110011...
Stop at any finite number of bits, and you get an approximation.
On a typical machine running Python, there are 53 bits of precision available
for a Python float, so the value stored internally when you enter the decimal
number 0.1 is the binary fraction
0.00011001100110011001100110011001100110011001100110011010
which is close to, but not exactly equal to, 1/10.
It’s easy to forget that the stored value is an approximation to the original decimal fraction, because of the way that floats are displayed at the interpreter prompt. Python only prints a decimal approximation to the true decimal value of the binary approximation stored by the machine. If Python were to print the true decimal value of the binary approximation stored for 0.1, it would have to display
>>> 0.1
0.1000000000000000055511151231257827021181583404541015625
这比大多数人认为有用的数字更多，因此Python通过显示舍入值来保持可管理的位数
>>> 0.1
0.1
It’s important to realize that this is, in a real sense, an illusion: the value in the machine is not exactly 1/10, you’re simply rounding the display of the true machine value. This fact becomes apparent as soon as you try to do arithmetic with these values
>>> 0.1 + 0.2
0.30000000000000004
请注意这种情况是二进制浮点数的本质特性：它不是 Python 的错误，也不是你代码中的错误。 你会在所有支持你的硬件中的浮点运算的语言中发现同样的情况（虽然某些语言在默认状态或所有输出模块下都不会 显示 这种差异）。
Other surprises follow from this one. For example, if you try to round the value 2.675 to two decimal places, you get this
>>> round(2.675, 2)
2.67
The documentation for the built-in round() function says that it rounds
to the nearest value, rounding ties away from zero. Since the decimal fraction
2.675 is exactly halfway between 2.67 and 2.68, you might expect the result
here to be (a binary approximation to) 2.68. It’s not, because when the
decimal string 2.675 is converted to a binary floating-point number, it’s
again replaced with a binary approximation, whose exact value is
2.67499999999999982236431605997495353221893310546875
Since this approximation is slightly closer to 2.67 than to 2.68, it’s rounded down.
If you’re in a situation where you care which way your decimal halfway-cases
are rounded, you should consider using the decimal module.
Incidentally, the decimal module also provides a nice way to “see” the
exact value that’s stored in any particular Python float
>>> from decimal import Decimal
>>> Decimal(2.675)
Decimal('2.67499999999999982236431605997495353221893310546875')
Another consequence is that since 0.1 is not exactly 1/10, summing ten values of 0.1 may not yield exactly 1.0, either:
>>> sum = 0.0
>>> for i in range(10):
... sum += 0.1
...
>>> sum
0.9999999999999999
二进制浮点运算会造成许多这样的“意外”。 有关 “0.1” 的问题会在下面的“表示性错误”一节中更详细地描述。 请参阅 浮点数的危险性 一文了解有关其他常见意外现象的更详细介绍。
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.
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. For
fine control over how a float is displayed see the str.format() method’s
format specifiers in Format String Syntax.
14.1. 表示性错误¶
本小节将详细解释 “0.1” 的例子，并说明你可以怎样亲自对此类情况进行精确分析。 假定前提是已基本熟悉二进制浮点表示法。
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:
>>> 0.1 + 0.2
0.30000000000000004
Why is that? 1/10 and 2/10 are not exactly representable as a binary fraction. Almost all machines today (July 2010) 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)
写为
J ~= 2**N / 10
并且由于 J 恰好有 53 位 (即 >= 2**52 但 < 2**53)，N 的最佳值为 56:
>>> 2**52
4503599627370496
>>> 2**53
9007199254740992
>>> 2**56/10
7205759403792793
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
由于余数超过 10 的一半，最佳近似值可通过四舍五入获得:
>>> q+1
7205759403792794
Therefore the best possible approximation to 1/10 in 754 double precision is that over 2**56, or
7205759403792794 / 72057594037927936
请注意由于我们做了向上舍入，这个结果实际上略大于 1/10；如果我们没有向上舍入，则商将会略小于 1/10。 但无论如何它都不会是 精确的 1/10！
因此计算永远不会“看到”1/10：它实际看到的就是上面所给出的小数，它所能达到的最佳 754 双精度近似值:
>>> .1 * 2**56
7205759403792794.0
If we multiply that fraction by 10**30, we can see the (truncated) value of its 30 most significant decimal digits:
>>> 7205759403792794 * 10**30 // 2**56
100000000000000005551115123125L
meaning that the exact number stored in the computer is approximately equal to the decimal value 0.100000000000000005551115123125. In versions prior to Python 2.7 and Python 3.1, Python rounded this value to 17 significant digits, giving ‘0.10000000000000001’. In current versions, Python displays a value based on the shortest decimal fraction that rounds correctly back to the true binary value, resulting simply in ‘0.1’.