将c++ DTW代码转换为java

converting c++ DTW code to java

本文关键字：java 转换 代码 c++ DTW      更新时间：2023-10-16

我想把一个代码从c++翻译成Java。原始代码实现了快速DTW算法。我弄不清楚的代码是I属性，我不确定它做什么，因此，我不能转换它。

Java中的错误是在语句l_buff+I &u_buff+I，因为int I &amp之间不支持加号;double[] l_buff,u_buff .

我已经包含了所有涉及I的语句

int  I;
for(i=0; i<ep; i++)
{
    /// A bunch of data has been read and pick one of them at a time to use
    d = buffer[i];
    /// Calculate sum and sum square
    ex += d;
    ex2 += d*d;
    /// t is a circular array for keeping current data
    t[i%m] = d;
    /// Double the size for avoiding using modulo "%" operator
    t[(i%m)+m] = d;
    /// Start the task when there are more than m-1 points in the current chunk
    if( i >= m-1 )
    {
        mean = ex/m;
        std = ex2/m;
        std = Math.sqrt(std-mean*mean);
        /// compute the start location of the data in the current circular array, t
        j = (i+1)%m;
        /// the start location of the data in the current chunk
        I = i-(m-1);
        lb_k2 = lb_keogh_data_cumulative(order, tz, qo, cb2, l_buff+I, u_buff+I, m, mean, std, bsf);

和lb_data_cumlative方法的实现是

public static double lb_keogh_data_cumulative(int[] order, double []tz, double []qo, double []cb, double []l, double []u, int len, double mean, double std, double best_so_far )
{
    double lb = 0;
    double uu,ll,d;
    for (int i = 0; i < len && lb < best_so_far; i++)
    {
        uu = (u[order[i]]-mean)/std;
        ll = (l[order[i]]-mean)/std;
        d = 0;
        if (qo[i] > uu)
            d = dist(qo[i], uu);
        else
        {
            if(qo[i] < ll)
                d = dist(qo[i], ll);
        }
        lb += d;
        cb[order[i]] = d;
    }
    return lb;
}

这里是代码所依赖的文件SIGKDD TRILLION

l_buff+I和u_buff+I表示将数组的开头移到I元素。lb_keogh_data_cumulative参数l和u不会看到给定数组的第一个I元素。

可以写成

lb_k2 = lb_keogh_data_cumulative(order, tz, qo, cb2, Arrays.copyOfRange(l_buff, I, l_buff.length), Arrays.copyOfRange(u_buff, I, u_buff.length), m, mean, std, bsf);

数组不会被调用的方法修改，所以你可以传递一个副本。

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