| .. _tut-fp-issues: | |
| ************************************************** | |
| Floating Point Arithmetic: Issues and Limitations | |
| ************************************************** | |
| .. sectionauthor:: Tim Peters <tim_one@users.sourceforge.net> | |
| 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. | |
| 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. | |
| 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. | |
| 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 :: | |
| 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 | |
| That is more digits than most people find useful, so Python keeps the number | |
| of digits manageable by displaying a rounded value instead :: | |
| >>> 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 | |
| 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). | |
| 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 :func:`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 :mod:`decimal` module. | |
| Incidentally, the :mod:`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 | |
| 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 <http://www.lahey.com/float.htm>`_ | |
| for a more complete account of other common surprises. | |
| 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 :meth:`str.format` method's | |
| format specifiers in :ref:`formatstrings`. | |
| .. _tut-fp-error: | |
| Representation Error | |
| ==================== | |
| 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. | |
| :dfn:`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) | |
| 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 | |
| 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 | |
| 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 | |
| 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:: | |
| >>> .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'. |