7

在 C# 中,我想将双精度数舍入到较低的精度,以便可以将它们存储在关联数组中不同大小的存储桶中。与通常的四舍五入不同,我想四舍五入到一些有效位。因此,大数的绝对值比小数的变化要大得多,但它们往往会按相同的比例发生变化。因此,如果我想四舍五入到 10 个二进制数字,我会找到 10 个最高有效位,并将所有低位清零,可能会添加一个小数字进行四舍五入。

我更喜欢四舍五入的“中途”数字。

如果它是整数类型,这将是一个可能的算法:

  1. Find: zero-based index of the most significant binary digit set H.
  2. Compute: B = H - P, 
       where P is the number of significant digits of precision to round
       and B is the binary digit to start rounding, where B = 0 is the ones place, 
       B = 1 is the twos place, etc. 
  3. Add: x = x + 2^B 
       This will force a carry if necessary (we round halfway values up).
  4. Zero out: x = x mod 2^(B+1). 
       This clears the B place and all lower digits.

问题是找到一种有效的方法来找到最高位集。如果我使用整数,有很酷的技巧可以找到 MSB。如果可以的话,我不想打电话给 Round(Log2(x)) 。该函数将被调用数百万次。

注意:我已经阅读了这个 SO 问题:

将双精度值舍入到(有点)较低精度的好方法是什么?

它适用于 C++。我正在使用 C#。

更新:

这是我正在使用的代码(根据回答者提供的内容修改):

/// <summary>
/// Round numbers to a specified number of significant binary digits.
/// 
/// For example, to 3 places, numbers from zero to seven are unchanged, because they only require 3 binary digits,
/// but larger numbers lose precision:
/// 
///      8    1000 => 1000   8
///      9    1001 => 1010  10
///     10    1010 => 1010  10
///     11    1011 => 1100  12
///     12    1100 => 1100  12
///     13    1101 => 1110  14
///     14    1110 => 1110  14
///     15    1111 =>10000  16
///     16   10000 =>10000  16
///     
/// This is different from rounding in that we are specifying the place where rounding occurs as the distance to the right
/// in binary digits from the highest bit set, not the distance to the left from the zero bit.
/// </summary>
/// <param name="d">Number to be rounded.</param>
/// <param name="digits">Number of binary digits of precision to preserve. </param>
public static double AdjustPrecision(this double d, int digits)
{
    // TODO: Not sure if this will work for both normalized and denormalized doubles. Needs more research.
    var shift = 53 - digits; // IEEE 754 doubles have 53 bits of significand, but one bit is "implied" and not stored.
    ulong significandMask = (0xffffffffffffffffUL >> shift) << shift;
    var local_d = d;
    unsafe
    {
        // double -> fixed point (sorta)
        ulong toLong = *(ulong*)(&local_d);
        // mask off your least-sig bits
        var modLong = toLong & significandMask;
        // fixed point -> float (sorta)
        local_d = *(double*)(&modLong);
    }
    return local_d;
}

更新 2:Dekker 算法

感谢另一位受访者,我从 Dekker 的算法中得出了这一点。它舍入到最接近的值,而不是像上面的代码那样截断,它只使用安全代码:

private static double[] PowersOfTwoPlusOne;

static NumericalAlgorithms()
{
    PowersOfTwoPlusOne = new double[54];
    for (var i = 0; i < PowersOfTwoPlusOne.Length; i++)
    {
        if (i == 0)
            PowersOfTwoPlusOne[i] = 1; // Special case.
        else
        {
            long two_to_i_plus_one = (1L << i) + 1L;
            PowersOfTwoPlusOne[i] = (double)two_to_i_plus_one;
        }
    }
}

public static double AdjustPrecisionSafely(this double d, int digits)
{
    double t = d * PowersOfTwoPlusOne[53 - digits];
    double adjusted = t - (t - d);
    return adjusted;
}

更新 2:时间

我跑了一个测试,发现 Dekker 的算法比 TWICE 快!

测试中的调用次数:100,000,000
不安全时间 = 1.922(秒)
安全时间 = 0.799(秒)

4

2 回答 2

8

Dekker 算法将浮点数拆分为高位和低位部分。如果有效数字中有s位(IEEE 754 64 位二进制中的 53 位),则*x0接收高s - b位,这是您请求的,并*x1接收剩余位,您可以丢弃这些位。在下面的代码中,Scale应该有值 2 b。如果b在编译时已知,例如常量 43,则可以替换Scale0x1p43. 否则,您必须以某种方式产生 2 b

这需要四舍五入到最近的模式。IEEE 754 算术就足够了,但其他合理的算术也可能没问题。它将关系四舍五入,这不是您所要求的(向上关系)。那有必要吗?

这假设x * (Scale + 1)不会溢出。必须以双精度(不更高)对操作进行评估。

void Split(double *x0, double *x1, double x)
{
    double d = x * (Scale + 1);
    double t = d - x;
    *x0 = d - t;
    *x1 = x - *x0;
}
于 2013-01-11T20:02:11.037 回答
2

有趣...从未听说过需要这样做,但我认为您可以通过一些时髦的不安全代码“做到”...

void Main()
{
    // how many bits you want "saved"
    var maxBits = 20;

    // create a mask like 0x1111000 where # of 1's == maxBits
    var shift = (sizeof(int) * 8) - maxBits;
    var maxBitsMask = (0xffffffff >> shift) << shift;

    // some floats
    var floats = new []{ 1.04125f, 2.19412347f, 3.1415926f};
    foreach (var f in floats)
    {
        var localf = f;
        unsafe
        {
            // float -> fixed point (sorta)
            int toInt = *(int*)(&localf);
            // mask off your least-sig bits
            var modInt = toInt & maxBitsMask;
            // fixed point -> float (sorta)
            localf = *(float*)(&modInt);
        }
        Console.WriteLine("Was {0}, now {1}", f, localf);
    }
}

和双打:

void Main()
{
    var maxBits = 50;
    var shift = (sizeof(long) * 8) - maxBits;
    var maxBitsMask = (0xffffffffffffffff >> shift) << shift;
    var doubles = new []{ 1412.04125, 22.19412347, 3.1415926};
    foreach (var d in doubles)
    {
        var local = d;
        unsafe
        {
            var toLong = *(ulong*)(&local);
            var modLong = toLong & maxBitsMask;
            local = *(double*)(&modLong);
        }
        Console.WriteLine("Was {0}, now {1}", d, local);
    }
}

哇...我不被接受。:)

为了完整起见,这里使用 Jeppe 的“不安全”方法:

void Main()
{
    var maxBits = 50;
    var shift = (sizeof(long) * 8) - maxBits;
    var maxBitsMask = (long)((0xffffffffffffffff >> shift) << shift);
    var doubles = new []{ 1412.04125, 22.19412347, 3.1415926};
    foreach (var d in doubles)
    {
        var local = d;
        var asLong = BitConverter.DoubleToInt64Bits(d);
        var modLong = asLong & maxBitsMask;
        local = BitConverter.Int64BitsToDouble(modLong);
        Console.WriteLine("Was {0}, now {1}", d, local);
    }
}
于 2013-01-11T20:10:15.780 回答