如何在.net舍System.Decimal到若干显著数字 [英] How to round System.Decimal in .Net to a number of significant figures

问题描述

我有一个System.Decimal数字

I have a System.Decimal number

0.00123456789

0.00123456789

，我希望舍入为3个有效数字中。我期望

and I wish to round to 3 significant figures. I expect

0.00123

，其行为是舍入行为，而不是截断。

with the behaviour to be a rounding behaviour rather than truncation. Is there a bullet proof way to do this in .Net?

推荐答案

你可以试试这个...但我不是保证任何...在20分钟内编写和测试，基于Pyrolistical的代码，来自 http://stackoverflow.com/a/1581007/613130有一个很大的区别，他使用 long 为移位变量因为 double 的精度为15-16位数，而 long 有18-19，因此 long 就足够了），而我使用 decimal （因为具有28-29位数的精度。）

You can try this... But I don't guarantee anything... Written and tested in 20 minutes and based on Pyrolistical's code from http://stackoverflow.com/a/1581007/613130 There is a big difference in that he uses a long for the shifted variable (because a double has a precision of 15-16 digits, while a long has 18-19, so a long is enough), while I use a decimal (because decimal has a precision of 28-29 digits).

public static decimal RoundToSignificantFigures(decimal num, int n)
{
    if (num == 0)
    {
        return 0;
    }

    // We are only looking for the next power of 10... 
    // The double conversion could impact in some corner cases,
    // but I'm not able to construct them...
    int d = (int)Math.Ceiling(Math.Log10((double)Math.Abs(num)));
    int power = n - d;

    // Same here, Math.Pow(10, *) is an integer number
    decimal magnitude = (decimal)Math.Pow(10, power);

    // I'm using the MidpointRounding.AwayFromZero . I'm not sure
    // having a MidpointRounding.ToEven would be useful (is Banker's
    // rounding used for significant figures?)
    decimal shifted = Math.Round(num * magnitude, 0, MidpointRounding.AwayFromZero);
    decimal ret = shifted / magnitude;

    return num >= 0 ? ret : -ret;
}

如果不信任 Math.Ceiling（Math.Log10（（double）可以使用：

private static readonly decimal[] Pows = Enumerable.Range(-28, 57)
    .Select(p => (decimal)Math.Pow(10, p))
    .ToArray();

public static int Log10Ceiling(decimal num)
{
    int log10 = Array.BinarySearch(Pows, num);
    return (log10 >= 0 ? log10 : ~log10) - 28;
}



我已经在另外20分钟已经测试了所有的 Math.Pow（（double），p）所有的值-28 - +28）。似乎工作，它只有20％基于 double s的C＃公式，它基于pows的静态数组和 BinarySearch 。幸运的是， code> BinarySearch 已经建议下一个元素，当它找不到一个:-)，因此 Ceiling 是免费的。

I have written it in another 20 minutes (and yes, I have tested all the Math.Pow((double), p) for all the values -28 - +28). It seems to work, and it's only 20% slower than the C# formula based on doubles). It's based on a static array of pows and a BinarySearch. Luckily the BinarySearch already "suggests" the next element when it can't find one :-), so the Ceiling is for free.

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