我想将代码从C ++转换为Java。原始代码实现了快速DTW算法。我不知道的那段代码是I
属性,我不确定它会做什么,因此我无法将其转换。
Java中的错误出现在语句l_buff+I
&中,u_buff+I
因为int I
&之间不支持plus运算符double[] l_buff,u_buff
。
我已经包括了所有涉及的陈述 I
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
方法实现
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;
}
这是代码依赖SIGKDD TRILLION的论文
l_buff+I
并u_buff+I
意味着你转移阵列的开始I
元素。的lb_keogh_data_cumulative
参数l
和u
将不会看到第一I
给定阵列的元件。
所以你可以写一些像
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,因此您可以传递副本。
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我来说两句