2007-08-15 14:28:22 +00:00
										 
									 
								 
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								.. _tut-fp-issues:
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								**************************************************
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								Floating Point Arithmetic:  Issues and Limitations
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								**************************************************
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								.. sectionauthor:: Tim Peters <tim_one@users.sourceforge.net>
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								Floating-point numbers are represented in computer hardware as base 2 (binary)
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								fractions.  For example, the decimal fraction ::
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								   0.125
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								has value 1/10 + 2/100 + 5/1000, and in the same way the binary fraction ::
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								   0.001
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								has value 0/2 + 0/4 + 1/8.  These two fractions have identical values, the only
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								real difference being that the first is written in base 10 fractional notation,
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								and the second in base 2.
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								Unfortunately, most decimal fractions cannot be represented exactly as binary
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								fractions.  A consequence is that, in general, the decimal floating-point
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								numbers you enter are only approximated by the binary floating-point numbers
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								actually stored in the machine.
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								The problem is easier to understand at first in base 10.  Consider the fraction
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								1/3.  You can approximate that as a base 10 fraction::
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								   0.3
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								or, better, ::
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								   0.33
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								or, better, ::
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								   0.333
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								and so on.  No matter how many digits you're willing to write down, the result
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								will never be exactly 1/3, but will be an increasingly better approximation of
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								1/3.
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								In the same way, no matter how many base 2 digits you're willing to use, the
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								decimal value 0.1 cannot be represented exactly as a base 2 fraction.  In base
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								2, 1/10 is the infinitely repeating fraction ::
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								   0.0001100110011001100110011001100110011001100110011...
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											2009-04-24 03:09:06 +00:00
										 
									 
								 
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								Stop at any finite number of bits, and you get an approximation.  On most
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								machines today, floats are approximated using a binary fraction with
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								the numerator using the first 53 bits following the most significant bit and
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								with the denominator as a power of two.  In the case of 1/10, the binary fraction
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								is ``3602879701896397 / 2 ** 55`` which is close to but not exactly
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								equal to the true value of 1/10.
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								Many users are not aware of the approximation because of the way values are
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								displayed.  Python only prints a decimal approximation to the true decimal
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								value of the binary approximation stored by the machine.  On most machines, if
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								Python were to print the true decimal value of the binary approximation stored
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								for 0.1, it would have to display ::
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								   >>> 0.1
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								   0.1000000000000000055511151231257827021181583404541015625
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											2007-08-15 14:28:22 +00:00
										 
									 
								 
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								That is more digits than most people find useful, so Python keeps the number
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								of digits manageable by displaying a rounded value instead ::
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											2009-04-24 03:09:06 +00:00
										 
									 
								 
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								   >>> 1 / 10
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								   0.1
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								Just remember, even though the printed result looks like the exact value
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								of 1/10, the actual stored value is the nearest representable binary fraction.
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											2009-04-24 03:09:06 +00:00
										 
									 
								 
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								Interestingly, there are many different decimal numbers that share the same
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								nearest approximate binary fraction.  For example, the numbers ``0.1`` and
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								``0.10000000000000001`` and
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								``0.1000000000000000055511151231257827021181583404541015625`` are all
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								approximated by ``3602879701896397 / 2 ** 55``.  Since all of these decimal
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											2009-04-24 19:06:29 +00:00
										 
									 
								 
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								values share the same approximation, any one of them could be displayed
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								while still preserving the invariant ``eval(repr(x)) == x``.
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								Historically, the Python prompt and built-in :func:`repr` function would chose
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								the one with 17 significant digits, ``0.10000000000000001``, Starting with
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								Python 3.1, Python (on most systems) is now able to choose the shortest of
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								these and simply display ``0.1``.
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											2007-08-15 14:28:22 +00:00
										 
									 
								 
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								Note that this is in the very nature of binary floating-point: this is not a bug
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								in Python, and it is not a bug in your code either.  You'll see the same kind of
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								thing in all languages that support your hardware's floating-point arithmetic
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								(although some languages may not *display* the difference by default, or in all
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								output modes).
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											2009-02-21 20:59:32 +00:00
										 
									 
								 
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								Python's built-in :func:`str` function produces only 12 significant digits, and
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								you may wish to use that instead.  It's unusual for ``eval(str(x))`` to
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								reproduce *x*, but the output may be more pleasant to look at::
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											2009-04-24 03:09:06 +00:00
										 
									 
								 
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								   >>> str(math.pi)
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								   '3.14159265359'
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								   >>> repr(math.pi)
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								   '3.141592653589793'
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								   >>> format(math.pi, '.2f')
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								   '3.14'
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								It's important to realize that this is, in a real sense, an illusion: you're
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								simply rounding the *display* of the true machine value.
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											2009-04-26 20:10:50 +00:00
										 
									 
								 
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								One illusion may beget another.  For example, since 0.1 is not exactly 1/10,
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											2009-04-26 21:37:46 +00:00
										 
									 
								 
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								summing three values of 0.1 may not yield exactly 0.3, either::
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								   >>> .1 + .1 + .1 == .3
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								   False
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								Also, since the 0.1 cannot get any closer to the exact value of 1/10 and
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								0.3 cannot get any closer to the exact value of 3/10, then pre-rounding with
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								:func:`round` function cannot help::
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								   >>> round(.1, 1) + round(.1, 1) + round(.1, 1) == round(.3, 1)
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								   False
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								Though the numbers cannot be made closer to their intended exact values,
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								the :func:`round` function can be useful for post-rounding so that results
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								have inexact values that are comparable to one another::
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								    >>> round(.1 + .1 + .1, 1) == round(.3, 1)
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								    True
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								Binary floating-point arithmetic holds many surprises like this.  The problem
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								with "0.1" is explained in precise detail below, in the "Representation Error"
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								section.  See `The Perils of Floating Point <http://www.lahey.com/float.htm>`_
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								for a more complete account of other common surprises.
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								As that says near the end, "there are no easy answers."  Still, don't be unduly
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								wary of floating-point!  The errors in Python float operations are inherited
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								from the floating-point hardware, and on most machines are on the order of no
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								more than 1 part in 2\*\*53 per operation.  That's more than adequate for most
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								tasks, but you do need to keep in mind that it's not decimal arithmetic, and
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								that every float operation can suffer a new rounding error.
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								While pathological cases do exist, for most casual use of floating-point
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								arithmetic you'll see the result you expect in the end if you simply round the
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								display of your final results to the number of decimal digits you expect.
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											2008-05-26 01:03:56 +00:00
										 
									 
								 
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								:func:`str` usually suffices, and for finer control see the :meth:`str.format`
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								method's format specifiers in :ref:`formatstrings`.
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											2008-10-05 17:57:52 +00:00
										 
									 
								 
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								For use cases which require exact decimal representation, try using the
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								:mod:`decimal` module which implements decimal arithmetic suitable for
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								accounting applications and high-precision applications.
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								Another form of exact arithmetic is supported by the :mod:`fractions` module
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								which implements arithmetic based on rational numbers (so the numbers like
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								1/3 can be represented exactly).
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											2007-08-31 03:25:11 +00:00
										 
									 
								 
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								If you are a heavy user of floating point operations you should take a look
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								at the Numerical Python package and many other packages for mathematical and
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								statistical operations supplied by the SciPy project. See <http://scipy.org>.
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								Python provides tools that may help on those rare occasions when you really
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								*do* want to know the exact value of a float.  The
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								:meth:`float.as_integer_ratio` method expresses the value of a float as a
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								fraction::
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								   >>> x = 3.14159
							 | 
						
					
						
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							 | 
							
							
								   >>> x.as_integer_ratio()
							 | 
						
					
						
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								   (3537115888337719L, 1125899906842624L)
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								Since the ratio is exact, it can be used to losslessly recreate the
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								original value::
							 | 
						
					
						
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								    >>> x == 3537115888337719 / 1125899906842624
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								    True
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							 | 
							
							
								The :meth:`float.hex` method expresses a float in hexadecimal (base
							 | 
						
					
						
							| 
								
							 | 
							
								
							 | 
							
								
							 | 
							
							
								16), again giving the exact value stored by your computer::
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							 | 
							
								
							 | 
							
							
								   >>> x.hex()
							 | 
						
					
						
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							 | 
							
								
							 | 
							
								
							 | 
							
							
								   '0x1.921f9f01b866ep+1'
							 | 
						
					
						
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								This precise hexadecimal representation can be used to reconstruct
							 | 
						
					
						
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							 | 
							
								
							 | 
							
							
								the float value exactly::
							 | 
						
					
						
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								    >>> x == float.fromhex('0x1.921f9f01b866ep+1')
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								    True
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								Since the representation is exact, it is useful for reliably porting values
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							 | 
							
							
								across different versions of Python (platform independence) and exchanging
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								data with other languages that support the same format (such as Java and C99).
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											2009-04-26 22:01:46 +00:00
										 
									 
								 
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								Another helpful tool is the :func:`math.fsum` function which helps mitigate
							 | 
						
					
						
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							 | 
							
							
								loss-of-precision during summation.  It tracks "lost digits" as values are
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							 | 
							
								
							 | 
							
							
								added onto a running total.  That can make a difference in overall accuracy
							 | 
						
					
						
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								so that the errors do not accumulate to the point where they affect the
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								final total:
							 | 
						
					
						
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								   >>> sum([0.1] * 10) == 1.0
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								   False
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								   >>> math.fsum([0.1] * 10) == 1.0
							 | 
						
					
						
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								   True
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											2008-10-05 16:46:29 +00:00
										 
									 
								 
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											2007-08-15 14:28:22 +00:00
										 
									 
								 
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								.. _tut-fp-error:
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								Representation Error
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							 | 
							
							
								====================
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								This section explains the "0.1" example in detail, and shows how you can perform
							 | 
						
					
						
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							 | 
							
								
							 | 
							
							
								an exact analysis of cases like this yourself.  Basic familiarity with binary
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							 | 
							
							
								floating-point representation is assumed.
							 | 
						
					
						
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								:dfn:`Representation error` refers to the fact that some (most, actually)
							 | 
						
					
						
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								decimal fractions cannot be represented exactly as binary (base 2) fractions.
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							 | 
							
							
								This is the chief reason why Python (or Perl, C, C++, Java, Fortran, and many
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											2009-04-24 03:09:06 +00:00
										 
									 
								 
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								others) often won't display the exact decimal number you expect.
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											2007-08-15 14:28:22 +00:00
										 
									 
								 
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								Why is that?  1/10 is not exactly representable as a binary fraction. Almost all
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								machines today (November 2000) use IEEE-754 floating point arithmetic, and
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								almost all platforms map Python floats to IEEE-754 "double precision".  754
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								doubles contain 53 bits of precision, so on input the computer strives to
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												Merged revisions 69129-69131,69139-69140,69143,69154-69159,69169,69288-69289,69293,69297-69301,69348 via svnmerge from
svn+ssh://pythondev@svn.python.org/python/trunk
........
  r69129 | benjamin.peterson | 2009-01-30 19:42:55 -0600 (Fri, 30 Jan 2009) | 1 line
  check the errno in bad fd cases
........
  r69130 | andrew.kuchling | 2009-01-30 20:50:09 -0600 (Fri, 30 Jan 2009) | 1 line
  Add a section
........
  r69131 | andrew.kuchling | 2009-01-30 21:26:02 -0600 (Fri, 30 Jan 2009) | 1 line
  Text edits and markup fixes
........
  r69139 | mark.dickinson | 2009-01-31 10:44:04 -0600 (Sat, 31 Jan 2009) | 2 lines
  Add an extra test for long <-> float hash equivalence.
........
  r69140 | benjamin.peterson | 2009-01-31 10:52:03 -0600 (Sat, 31 Jan 2009) | 1 line
  PyErr_BadInternalCall() raises a SystemError, not TypeError #5112
........
  r69143 | benjamin.peterson | 2009-01-31 15:00:10 -0600 (Sat, 31 Jan 2009) | 1 line
  I believe the intention here was to avoid a global lookup
........
  r69154 | benjamin.peterson | 2009-01-31 16:33:02 -0600 (Sat, 31 Jan 2009) | 1 line
  fix indentation in comment
........
  r69155 | david.goodger | 2009-01-31 16:53:46 -0600 (Sat, 31 Jan 2009) | 1 line
  markup fix
........
  r69156 | gregory.p.smith | 2009-01-31 16:57:30 -0600 (Sat, 31 Jan 2009) | 4 lines
  - Issue #5104: The socket module now raises OverflowError when 16-bit port and
    protocol numbers are supplied outside the allowed 0-65536 range on bind()
    and getservbyport().
........
  r69157 | benjamin.peterson | 2009-01-31 17:43:25 -0600 (Sat, 31 Jan 2009) | 1 line
  add explanatory comment
........
  r69158 | benjamin.peterson | 2009-01-31 17:54:38 -0600 (Sat, 31 Jan 2009) | 1 line
  more flags which only work for function blocks
........
  r69159 | gregory.p.smith | 2009-01-31 18:16:01 -0600 (Sat, 31 Jan 2009) | 2 lines
  Update doc wording as suggested in issue4903.
........
  r69169 | guilherme.polo | 2009-01-31 20:56:16 -0600 (Sat, 31 Jan 2009) | 3 lines
  Restore Tkinter.Tk._loadtk so this test doesn't fail for problems
  related to ttk.
........
  r69288 | georg.brandl | 2009-02-05 04:30:57 -0600 (Thu, 05 Feb 2009) | 1 line
  #5153: fix typo in example.
........
  r69289 | georg.brandl | 2009-02-05 04:37:07 -0600 (Thu, 05 Feb 2009) | 1 line
  #5144: document that PySys_SetArgv prepends the script directory (or the empty string) to sys.path.
........
  r69293 | georg.brandl | 2009-02-05 04:59:28 -0600 (Thu, 05 Feb 2009) | 1 line
  #5059: fix example.
........
  r69297 | georg.brandl | 2009-02-05 05:32:18 -0600 (Thu, 05 Feb 2009) | 1 line
  #5015: document PythonHome API functions.
........
  r69298 | georg.brandl | 2009-02-05 05:33:21 -0600 (Thu, 05 Feb 2009) | 1 line
  #4827: fix callback example.
........
  r69299 | georg.brandl | 2009-02-05 05:35:28 -0600 (Thu, 05 Feb 2009) | 1 line
  #4820: use correct module for ctypes.util.
........
  r69300 | georg.brandl | 2009-02-05 05:38:23 -0600 (Thu, 05 Feb 2009) | 1 line
  #4563: disable alpha and roman lists, fixes wrong formatting of contributor list.
........
  r69301 | georg.brandl | 2009-02-05 05:40:35 -0600 (Thu, 05 Feb 2009) | 1 line
  #5031: fix Thread.daemon property docs.
........
  r69348 | benjamin.peterson | 2009-02-05 19:47:31 -0600 (Thu, 05 Feb 2009) | 1 line
  fix download link
........
											
										 
										
											2009-02-06 02:40:07 +00:00
										 
									 
								 
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								convert 0.1 to the closest fraction it can of the form *J*/2**\ *N* where *J* is
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											2007-08-15 14:28:22 +00:00
										 
									 
								 
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								an integer containing exactly 53 bits.  Rewriting ::
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								   1 / 10 ~= J / (2**N)
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								as ::
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								   J ~= 2**N / 10
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								and recalling that *J* has exactly 53 bits (is ``>= 2**52`` but ``< 2**53``),
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								the best value for *N* is 56::
							 | 
						
					
						
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								   >>> 2**52
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											2008-08-10 12:16:45 +00:00
										 
									 
								 
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								   4503599627370496
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											2007-08-15 14:28:22 +00:00
										 
									 
								 
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								   >>> 2**53
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											2008-08-10 12:16:45 +00:00
										 
									 
								 
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								   9007199254740992
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											2007-08-15 14:28:22 +00:00
										 
									 
								 
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								   >>> 2**56/10
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											2008-08-10 12:16:45 +00:00
										 
									 
								 
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							 | 
							
							
								   7205759403792794.0
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											2007-08-15 14:28:22 +00:00
										 
									 
								 
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								That is, 56 is the only value for *N* that leaves *J* with exactly 53 bits.  The
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								best possible value for *J* is then that quotient rounded::
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								   >>> q, r = divmod(2**56, 10)
							 | 
						
					
						
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								   >>> r
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											2008-08-10 12:16:45 +00:00
										 
									 
								 
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								   6
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											2007-08-15 14:28:22 +00:00
										 
									 
								 
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								Since the remainder is more than half of 10, the best approximation is obtained
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								by rounding up::
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							 | 
						
					
						
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								   >>> q+1
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											2008-08-10 12:16:45 +00:00
										 
									 
								 
							 | 
							
								
									
										
									
								
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								   7205759403792794
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											2007-08-15 14:28:22 +00:00
										 
									 
								 
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								Therefore the best possible approximation to 1/10 in 754 double precision is
							 | 
						
					
						
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							 | 
							
								
							 | 
							
							
								that over 2\*\*56, or ::
							 | 
						
					
						
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								   7205759403792794 / 72057594037927936
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											2009-04-24 03:09:06 +00:00
										 
									 
								 
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								Dividing both the numerator and denominator by two reduces the fraction to::
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								   3602879701896397 / 36028797018963968
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											2007-08-15 14:28:22 +00:00
										 
									 
								 
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								Note that since we rounded up, this is actually a little bit larger than 1/10;
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								if we had not rounded up, the quotient would have been a little bit smaller than
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								1/10.  But in no case can it be *exactly* 1/10!
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								So the computer never "sees" 1/10:  what it sees is the exact fraction given
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								above, the best 754 double approximation it can get::
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											2009-04-24 03:09:06 +00:00
										 
									 
								 
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								   >>> 0.1 * 2 ** 55
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								   3602879701896397.0
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											2009-04-24 03:09:06 +00:00
										 
									 
								 
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								If we multiply that fraction by 10\*\*60, we can see the value of out to
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								60 decimal digits::
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											2009-04-24 03:09:06 +00:00
										 
									 
								 
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								   >>> 3602879701896397 * 10 ** 60 // 2 ** 55
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								   1000000000000000055511151231257827021181583404541015625
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								meaning that the exact number stored in the computer is approximately equal to
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								the decimal value 0.100000000000000005551115123125.  Rounding that to 17
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								significant digits gives the 0.10000000000000001 that Python displays (well,
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								will display on any 754-conforming platform that does best-possible input and
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								output conversions in its C library --- yours may not!).
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											2009-04-24 03:09:06 +00:00
										 
									 
								 
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								The :mod:`fractions` and :mod:`decimal` modules make these calculations
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								easy::
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											2007-08-15 14:28:22 +00:00
										 
									 
								 
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											2009-04-24 03:09:06 +00:00
										 
									 
								 
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								   >>> from decimal import Decimal
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								   >>> from fractions import Fraction
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								   >>> print(Fraction.from_float(0.1))
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								   3602879701896397/36028797018963968
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								   >>> print(Decimal.from_float(0.1))
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								   0.1000000000000000055511151231257827021181583404541015625
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