Julia中的精确十进制算术 [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 中.我想以 CPU 高效的方式获得 0.16 的数学正确答案.我知道 round() 有效，但这需要事先了解答案占据的小数位数，因此它不是通用解决方案.

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.

推荐答案

一些选项:

1. 使用内置的 Rational 类型.最准确和最快的方法是

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

16//100 * 16//100

16//100 * 16//100

如果你使用非常大的数字，这些可能会溢出，在这种情况下你可以使用 BigInts 代替，

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)

(您实际上不需要将它们全部包装在 big 中，因为有理数会自动提升).

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

您也可以使用 rationalize(0.16)，但这可能不够准确或高效，因为文字 0.16 已经转换为 Float64 到 Julia 看到它时，所以您正在转换为二进制浮点，然后转换为 Rational.

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.

1. DecFP.jl 封装了 Intel 的 IEEE-754 十进制浮点数.这应该相当快(尽管不如二进制效率高)，但具有固定的精度，因此您必须在某些时候进行舍入.

1. 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.

Decimals.jl 是一个大十进制"浮点库:如它使用任意精度算法，它会比 DecFP 慢.

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

要说哪个是最好的，需要更多关于您的预期用途的信息.

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

这篇关于Julia中的精确十进制算术的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！