# 15. 浮点算术：争议和限制¶

Floating-point numbers are represented in computer hardware as base 2 (binary) fractions. For example, the decimal fraction `0.625` has value 6/10 + 2/100 + 5/1000, and in the same way the binary fraction `0.101` has value 1/2 + 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.

```0.3
```

```0.33
```

```0.333
```

```0.0001100110011001100110011001100110011001100110011...
```

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:

```>>> 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:

```>>> 1 / 10
0.1
```

For more pleasant output, you may wish to use string formatting to produce a limited number of significant digits:

```>>> format(math.pi, '.12g')  # give 12 significant digits
'3.14159265359'

>>> format(math.pi, '.2f')   # give 2 digits after the point
'3.14'

>>> repr(math.pi)
'3.141592653589793'
```

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:

```>>> 0.1 + 0.1 + 0.1 == 0.3
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:

```>>> round(0.1, 1) + round(0.1, 1) + round(0.1, 1) == round(0.3, 1)
False
```

Though the numbers cannot be made closer to their intended exact values, the `math.isclose()` function can be useful for comparing inexact values:

```>>> math.isclose(0.1 + 0.1 + 0.1, 0.3)
True
```

Alternatively, the `round()` function can be used to compare rough approximations:

```.. doctest::
```
```>>> round(math.pi, ndigits=2) == round(22 / 7, ndigits=2)
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 Examples of Floating Point Problems for a pleasant summary of how binary floating-point works and the kinds of problems commonly encountered in practice. Also see The Perils of Floating Point for a more complete account of other common surprises.

If you are a heavy user of floating-point operations you should take a look at the NumPy package and many other packages for mathematical and statistical operations supplied by the SciPy project. See <https://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:

```>>> x = 3.14159
>>> x.as_integer_ratio()
(3537115888337719, 1125899906842624)
```

Since the ratio is exact, it can be used to losslessly recreate the original value:

```>>> x == 3537115888337719 / 1125899906842624
True
```

The `float.hex()` method expresses a float in hexadecimal (base 16), again giving the exact value stored by your computer:

```>>> x.hex()
'0x1.921f9f01b866ep+1'
```

This precise hexadecimal representation can be used to reconstruct the float value exactly:

```>>> x == float.fromhex('0x1.921f9f01b866ep+1')
True
```

Another helpful tool is the `sum()` function which helps mitigate loss-of-precision during summation. It uses extended precision for intermediate rounding steps 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:

```>>> 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 == 1.0
False
>>> sum([0.1] * 10) == 1.0
True
```

The `math.fsum()` goes further and tracks all of the "lost digits" as values are added onto a running total so that the result has only a single rounding. This is slower than `sum()` but will be more accurate in uncommon cases where large magnitude inputs mostly cancel each other out leaving a final sum near zero:

```>>> arr = [-0.10430216751806065, -266310978.67179024, 143401161448607.16,
...        -143401161400469.7, 266262841.31058735, -0.003244936839808227]
>>> float(sum(map(Fraction, arr)))   # Exact summation with single rounding
8.042173697819788e-13
>>> math.fsum(arr)                   # Single rounding
8.042173697819788e-13
>>> sum(arr)                         # Multiple roundings in extended precision
8.042178034628478e-13
>>> total = 0.0
>>> for x in arr:
...     total += x                   # Multiple roundings in standard precision
...
>>> total                            # Straight addition has no correct digits!
-0.0051575902860057365
```

## 15.1. 表示性错误¶

Why is that? 1/10 is not exactly representable as a binary fraction. Since at least 2000, almost all machines use IEEE 754 binary floating-point arithmetic, and almost all platforms map Python floats to IEEE 754 binary64 "double precision" values. IEEE 754 binary64 values 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)
```

```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 <=  2**56 // 10  < 2**53
True
```

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 IEEE 754 double precision is:

```7205759403792794 / 2 ** 56
```

```3602879701896397 / 2 ** 55
```

So the computer never "sees" 1/10: what it sees is the exact fraction given above, the best IEEE 754 double approximation it can get:

```>>> 0.1 * 2 ** 55
3602879701896397.0
```

If we multiply that fraction by 10**55, we can see the value out to 55 decimal digits:

```>>> 3602879701896397 * 10 ** 55 // 2 ** 55
1000000000000000055511151231257827021181583404541015625
```

meaning that the exact number stored in the computer is equal to the decimal value 0.1000000000000000055511151231257827021181583404541015625. Instead of displaying the full decimal value, many languages (including older versions of Python), round the result to 17 significant digits:

```>>> format(0.1, '.17f')
'0.10000000000000001'
```

The `fractions` and `decimal` modules make these calculations easy:

```>>> from decimal import Decimal
>>> from fractions import Fraction

>>> Fraction.from_float(0.1)
Fraction(3602879701896397, 36028797018963968)

>>> (0.1).as_integer_ratio()
(3602879701896397, 36028797018963968)

>>> Decimal.from_float(0.1)
Decimal('0.1000000000000000055511151231257827021181583404541015625')

>>> format(Decimal.from_float(0.1), '.17')
'0.10000000000000001'
```