在Julia中精确的十进制算术 [英] Exact decimal arithmetic in Julia

问题描述

由于浮点数学的性质， .4 * .4 = 0.16000000000000003 在Julia。我想以CPU高效的方式得到 0.16 的数学正确答案。我知道round（）的作品，但这需要预先知道小数位数的答案占据，所以这不是一个普遍的解决方案。 解决方案

一些选项：
$ b

1. 使用内置的 Rational 类型。最准确和最快捷的方式是
   $ b $ 16 // 100 * 16 // 100

如果使用的数字非常大，可能会溢出，在这种情况下，您可以使用 BigInt s来替代，

$ $ p $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

（你实际上并不需要把它们都包装在 big 中，自动提升）。

您也可以使用 rationalize（0.16），但这可能不是那么准确或有效，在Julia看到它的时候，文字 0.16 已经被转换为 Float64 ，所以你要转换成二进制然后转换为 Rational 。

2. DecFP.jl 包装了英特尔实施的 IEEE-754 十进制浮点数。这应该是相当快（虽然不如二进制），但具有固定的精度，所以你将不得不在一定程度上舍入。

=https://github.com/tinybike/Decimals.jl =nofollow noreferrer> Decimals.jl 是一个十进制大数的浮点库：因为它使用任意的精度算法，要比DecFP慢。

要说哪一个是最好的，需要更多关于您的预期用途的信息。 p>

Due to the nature of floating-point math, .4 * .4 = 0.16000000000000003 in Julia. I want to get the mathematically correct answer of 0.16, in a CPU-efficient way. I know round() works, but that requires prior knowledge of the number of decimal places the answer occupies, so it isn't a general solution.

解决方案

Some options:

1. Use the inbuilt Rational type. The most accurate and fastest way would be

   16//100 * 16//100

If you're using very big numbers these might overflow, in which case you can use BigInts instead,

big(16)//big(100) * big(16)//big(100)

(you don't actually need to wrap them all in bigs, as the rationals will promote automatically).

You can also use rationalize(0.16), but this may not be quite as accurate or efficient, as the literal 0.16 has already been converted to a Float64 by the time Julia sees it, so you're converting to a binary floating point and then to a Rational.

2. DecFP.jl wraps the Intel implementation of IEEE-754 Decimal floating point. This should be reasonably fast (though not as efficient as binary), but has fixed precision, so you will have to round at some point.

3. Decimals.jl is a "big decimal" floating point library: as it uses arbitrary precision arithmetic, it is going to be slower than DecFP.

To say which is the best would require more information about your intended use.

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