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comparing float inequality (in python)
(2 answers)
Closed 9 years ago.
0.1 + 0.2 == 0.3 ==> False
I tried this in python, c#, c++, F#, Visual Basic.NET,ASP.NET!
0.1 + 0.2 == 0.30000000000000004 ==> True
This is True for all languages I mentioned above. Why this illogical inequality happens?
Python has a decimal library that will allow you to evaluate this to true (while also explaining that why its false as you have it), and in fact, they use almost the same exact example: http://docs.python.org/2/library/decimal.html
The exactness carries over into arithmetic. In decimal floating point, 0.1 + 0.1 + 0.1 - 0.3 is exactly equal to zero
Also related, "What every computer scientist should know about floating point", a seminal article written many years ago and still true today: http://www.fer.unizg.hr/_download/repository/paper%5B1%5D.pdf
You might want to give this a read: http://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html (What Every Computer Scientist Should Know About Floating-Point Arithmetic)
The short answer is that in binary 0.1 is a repeating fraction, it's "somewhere" between 1/8 and 1/16 so there is no 'bit exact' representation of 0.1 (or 0.2). When comparing floating point values you always have to do so within an epsilon value the prevent such problems.
Read this from the Python docs; it applies pretty much verbatim to all languages:
http://docs.python.org/2/tutorial/floatingpoint.html
For comparing floating point number, use this pattern:
Math.Abs(NumberToCompare1 - NumberToCompare2) < 0.01
Where 0.01 is the Epsilon, the precision of the operation.
Related
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Is floating point math broken?
(31 answers)
Closed 4 months ago.
I have a problem with conversation of double values.
In the picture Project's framework is .NET6 and i tried on .NET5 and i get same value again on x64.
Can you explain this situation to me?
And how can I get the value unchanged on x64?
Thank you.
This is expected behavior: floating point numbers have a limited number of significant digits, and a number that has a finite number of digits in their decimal representation may require infinite digits in their binary representation.
E.g., the decimal 0.1 is, in binary, 0.00011001100110011.... repeating. Storing this in a float or double, the number of digits is truncated.
Floating point numbers behave similar, but not identical to "real" numbers. This is something you should be aware of.
For financial mathematics, in C#, use the decimal type.
Here you find a good "what every developer should know" overview: https://floating-point-gui.de/.
The standard governing the most common (almost ubiquitous) implementation of floating point types is IEEE 754.
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c# string to float conversion invalid?
(3 answers)
Closed 5 years ago.
float.Parse("534818068")
returns: 534818080
I understand that there are many complications with float and decimal values. But maybe someone could explain this behaviour to me.
Thanks!
Floating point numbers have a relative precision, i.e. something like 7 or 8 digits. So only the first 7 or 8 digits are correct, independent of the actual total size of the number.
Floating point numbers are stored internally using the IEEE 754 standard (a sign, a biased exponent and a fraction).
float numbers are stored with a 32 bits representation, which means they will have a precision of 7 digits.
On the other hand, double are stored with a 64 bits representation, thus having 15-16 digits (source).
Which is why you shouldn't usually compare floats for equality for instance.
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Is floating point math broken?
(31 answers)
Closed 7 years ago.
AFAIK .NET's default round option is to even, so Math.Round(1.225,2) should give 1.22 but it gives 1.23.
Math.Round(2.225,2) = 2.22
Math.Round(100.225,2) = 100.22
all the values I tried rounds to nearest even but only 1.225 and -1.225 rounds to 1.23 and -1.23.
The main problem is that in float and double, the amount of decimal places is not part of the value, and the precision isn't decimal, but rather, binary. And there's no finite binary number that can represent 1.225 exactly.
So when you do Math.Round(1.225f, 2), you're actually doing something more like Math.Round(1.22500002f, 2) - there's no midpoint rounding involved.
The same problem appears with Math.Round(2.225f, 2) - it's just that the "real" value is slightly smaller than 2.225f, so the result rounds down. But there's still no midpoint rounding involved.
If you need decimal precision, use decimal. Neither float nor double are designed for decimal precision - they're fine for e.g. physics calculations, but not for e.g. accounting.
1.225 can't be represented exactly in floating point, so you're really rounding 1.225000023841858.
This question already has answers here:
C# - (int)Math.Round((double)(3514 + 3515)/2) =3514?
(2 answers)
Closed 9 years ago.
Similar question, but without double and 3 decimal places.
The difference is that the average of two integers we may have a double as a result, but when we use (int) Math.Ceiling ((double) value), result an integer.
C# - (int)Math.Round((double)(3514 + 3515)/2) =3514?
But in this case, we have two doubles and
Math.Round(((4.006+4.007)/2),3); // returns 4.006
Math.Round(((4.008+4.007)/2),3); // returns 4.008
WHY?
From the MSDN:
Return Value
Type: System.Double The integer nearest a. If the fractional component
of a is halfway between two integers, one of which is even and the
other odd, then the even number is returned. Note that this method
returns a Double instead of an integral type.
Remarks
The behavior of this method follows IEEE Standard 754, section 4. This
kind of rounding is sometimes called rounding to nearest, or banker's
rounding. It minimizes rounding errors that result from consistently
rounding a midpoint value in a single direction.
Also check this related thread
This question already has answers here:
Closed 12 years ago.
Possible Duplicate:
Why is floating point arithmetic in C# imprecise?
Console.WriteLine(0.5f * 2f); // 1
Console.WriteLine(0.5f * 2f - 1f); // 0
Console.WriteLine(0.1f * 10f); // 1
Console.WriteLine(0.1f * 10f - 1f); // 1.490116E-08
Why does 0.1f * 10f - 1f end up being 1.490116E-08 (0.0000001490116)?
See Wiki: Floating Point. float/double/decimal are relative precision types. Not all values (of which there are infinitely many) can be exactly stored. You are seeing the result of this loss of accuracy. This is why it is almost always correct to use |a - b| < small_delta for float-point comparisons.
Because floating operations are not precise, take a look here:
http://en.wikipedia.org/wiki/Floating_point
section Some other computer representations for non-integral numbers. 0.1 cannot finitely be represented in base 2.
Take a look at What Every Computer Scientist Should Know About Floating-Point Arithmetic.
Read this (explains quite exactly this case): http://www.exploringbinary.com/the-answer-is-one-unless-you-use-floating-point/
Easy, approximations accumulate.
0.5 is representable exactly in floating point, which is why 0.5*2 = 1 exactly.
However, 0.1 is not representable exactly, hence 0.1*10 is not exactly 1.