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:
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)