我必须存储几个非常低的概率值的产品(例如1E-80).使用原始的
java double将导致零,因为下溢.我不希望该值为零,因为稍后会有一个较大的数字(例如1E100),这将使值在双重可以处理的范围内.
所以,我自己创建了一个不同的类(MyDouble),它用于保存基本部分和指数部分.当进行计算时,例如乘法,我乘以基本部分,并添加指数.
该程序与原始双重类型很快.但是,当我使用我自己的类(MyDouble)时,程序真的很慢.我认为这是因为每次创建简单操作时都需要创建新的对象,而当不再需要对象时,垃圾回收器必须做很多工作.
我的问题是,有没有更好的方式,你认为我可以解决这个问题？如果没有,有没有办法,我可以用我自己的课程(MyDouble)来加速程序？
[注意：采取日志,后来取指数不能解决我的问题]
MyDouble类：
public class MyDouble {
public MyDouble(double base,int power){
this.base = base;
this.power = power;
}
public static MyDouble multiply(double... values) {
MyDouble returnMyDouble = new MyDouble(0);
double prodBase = 1;
int prodPower = 0;
for( double val : values) {
MyDouble ad = new MyDouble(val);
prodBase *= ad.base;
prodPower += ad.power;
}
String newBaseString = "" + prodBase;
String[] splitted = newBaseString.split("E");
double newBase = 0; int newPower = 0;
if(splitted.length == 2) {
newBase = Double.parseDouble(splitted[0]);
newPower = Integer.parseInt(splitted[1]);
} else {
newBase = Double.parseDouble(splitted[0]);
newPower = 0;
}
returnMyDouble.base = newBase;
returnMyDouble.power = newPower + prodPower;
return returnMyDouble;
}
}
解决方法
缓慢可能是因为在split和string concats中创建的中间字符串对象.
尝试这个：
/**
* value = base * 10 ^ power.
*/
public class MyDouble {
// Threshold values to determine whether given double is too small or not.
private static final double SMALL_EPSILON = 1e-8;
private static final double SMALL_EPSILON_MULTIPLIER = 1e8;
private static final int SMALL_EPSILON_POWER = 8;
private double myBase;
private int myPower;
public MyDouble(double base,int power){
myBase = base;
myPower = power;
}
public MyDouble(double base)
{
myBase = base;
myPower = 0;
adjustPower();
}
/**
* If base value is too small,increase the base by multiplying with some number and
* decrease the power accordingly.
* <p> E.g 0.000 000 000 001 * 10^1 => 0.0001 * 10^8
*/
private void adjustPower()
{
// Increase the base & decrease the power
// if given double value is less than threshold.
if (myBase < SMALL_EPSILON) {
myBase = myBase * SMALL_EPSILON_MULTIPLIER;
myPower -= SMALL_EPSILON_POWER;
}
}
/**
* This method multiplies given double and updates this object.
*/
public void multiply(MyDouble d)
{
myBase *= d.myBase;
myPower += d.myPower;
adjustPower();
}
/**
* This method multiplies given primitive double value with this object and update the
* base and power.
*/
public void multiply(double d)
{
multiply(new MyDouble(d));
}
@Override
public String toString()
{
return "Base:" + myBase + ",Power=" + myPower;
}
/**
* This method multiplies given double values and returns MyDouble object.
* It make sure that too small double values do not zero out the multiplication result.
*/
public static MyDouble multiply(double...values)
{
MyDouble result = new MyDouble(1);
for (int i=0; i<values.length; i++) {
result.multiply(values[i]);
}
return result;
}
public static void main(String[] args) {
MyDouble r = MyDouble.multiply(1e-80,1e100);
System.out.println(r);
}
}
如果您的目的仍然很慢,您可以修改multiply()方法直接对原始double进行操作,而不是创建一个MyDouble对象.