有没有办法让＆QUOT;显著数字＆QUOT;的小数？ [英] Is there a way to get the "significant figures" of a decimal?
问题描述
确定，经过一番调查，并在很大一部分要归功于乔恩和汉斯提供了有益的答案，这就是我能够放在一起。到目前为止，我觉得它似乎运作良好。我不会赌当然，我的生活就其总的正确性。
OK, after some investigation, and thanks in big part to the helpful answers provided by Jon and Hans, this is what I was able to put together. So far I think it seems to work well. I wouldn't bet my life on its total correctness, of course.
public static int GetSignificantDigitCount(this decimal value)
{
/* So, the decimal type is basically represented as a fraction of two
* integers: a numerator that can be anything, and a denominator that is
* some power of 10.
*
* For example, the following numbers are represented by
* the corresponding fractions:
*
* VALUE NUMERATOR DENOMINATOR
* 1 1 1
* 1.0 10 10
* 1.012 1012 1000
* 0.04 4 100
* 12.01 1201 100
*
* So basically, if the magnitude is greater than or equal to one,
* the number of digits is the number of digits in the numerator.
* If it's less than one, the number of digits is the number of digits
* in the denominator.
*/
int[] bits = decimal.GetBits(value);
if (value >= 1M || value <= -1M)
{
int highPart = bits[2];
int middlePart = bits[1];
int lowPart = bits[0];
decimal num = new decimal(lowPart, middlePart, highPart, false, 0);
int exponent = (int)Math.Ceiling(Math.Log10((double)num));
return exponent;
}
else
{
int scalePart = bits[3];
// Accoring to MSDN, the exponent is represented by
// bits 16-23 (the 2nd word):
// http://msdn.microsoft.com/en-us/library/system.decimal.getbits.aspx
int exponent = (scalePart & 0x00FF0000) >> 16;
return exponent + 1;
}
}
我还没有彻底测试这一切。这里有一些样品输入/输出，但：
I haven't tested it all that thoroughly. Here are a few sample inputs/outputs, though:
Value Precision
0 1 digit(s).
0.000 4 digit(s).
1.23 3 digit(s).
12.324 5 digit(s).
1.2300 5 digit(s).
-5 1 digit(s).
-5.01 3 digit(s).
-0.012 4 digit(s).
-0.100 4 digit(s).
0.0 2 digit(s).
10443.31 7 digit(s).
-130.340 6 digit(s).
-80.8000 6 digit(s).
使用这个code，我想我会实现我的目标做这样的事情：
Using this code, I imagine I would accomplish my goal by doing something like this:
public static decimal DivideUsingLesserPrecision(decimal x, decimal y)
{
int xDigitCount = x.GetSignificantDigitCount();
int yDigitCount = y.GetSignificantDigitCount();
int lesserPrecision = System.Math.Min(xDigitCount, yDigitCount);
return System.Math.Round(x / y, lesserPrecision);
}
我还没有真正完成过这样的工作，虽然。如果谁想要分享的想法：那会是多少AP preciated
I haven't really finished working through this, though. Anybody who wants to share thoughts: that would be much appreciated!
假设我有写这篇code：
Suppose I have write this code:
decimal a = 1.23M;
decimal b = 1.23000M;
Console.WriteLine(a);
Console.WriteLine(b);
以上将输出：
The above will output:
1.23
1.23000
我觉得这也适用，如果我用 decimal.Parse（1.23）为 A 和 decimal.Parse（1.23000）为 B （这意味着这个问题适用于在节目接收用户输入的情况下）
I find that this also works if I use decimal.Parse("1.23") for a and decimal.Parse("1.23000") for b (which means this question applies to cases where the program receives user input).
所以显然十进制值为某种方式知道什么，我会打电话给它的 precision 的。但是，我看到了十进制没有成员键入提供任何访问该的，除了的ToString 本身。
So clearly a decimal value is somehow "aware" of what I'll call its precision. However, I see no members on the decimal type that provide any way of accessing this, aside from ToString itself.
假设我想将两个十进制值和修剪的结果不太precise参数的precision。换句话说：
Suppose I wanted to multiply two decimal values and trim the result to the precision of the less precise argument. In other words:
decimal a = 123.4M;
decimal b = 5.6789M;
decimal x = a / b;
Console.WriteLine(x);
以上输出：
21.729560302171195125816619416
我要问的是：我怎么会写一个方法，将返回 21.73 ，而不是（因为 123.4M 有四个显著数字）？
What I'm asking is: how could I write a method that would return 21.73 instead (since 123.4M has four significant figures)?
需要明确的是：我知道我可以叫的ToString 的两个参数，每个字符串中算显著的数字，并用这个号圆计算的结果。我在寻找一个的不同的的方式，如果可能的话。
To be clear: I realize I could call ToString on both arguments, count the significant figures in each string, and use this number to round the result of the calculation. I'm looking for a different way, if possible.
（我的也的实现，在大多数情况下，您正在处理显著的数字，你可能并不需要使用的十进制类型。但是，我这么问是因为，正如我在一开始提到的，十进制键入显示的，包括约precision信息，而双没有，据我所知。）
(I also realize that in most scenarios where you're dealing with significant figures, you probably don't need to be using the decimal type. But I'm asking because, as I mentioned in the beginning, the decimal type appears to include information about precision, whereas double does not, as far as I know.)
推荐答案
您可以使用的 Decimal.GetBits ，以获得原始数据，并从工作了。
You can use Decimal.GetBits to get the raw data, and work it out from that.
不幸的是我没有时间在此刻写样本code - 你可能会想使用的 的BigInteger 的一些操作，如果您使用的是.NET 4 - 但希望这将让你去。刚参加工作了precision，然后调用 Math.Round 上的原始结果很可能是一个良好的开端。
Unfortunately I don't have time to write sample code at the moment - and you'll probably want to use BigInteger for some of the manipulation, if you're using .NET 4 - but hopefully this will get you going. Just working out the precision and then calling Math.Round on the original result may well be a good start.
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