c++ - 如何在 C++ 中快速计算 vector 的归一化 l1 和 l2 范数？

Question

我有一个矩阵 X，它在 d 维空间中有 n 列数据 vector 。给定一个 vector xj，v[j]是它的l1范数(所有abs(xji)的总和), w[j] 是它的 l2 范数的平方(所有 xji^2 的总和)，而 pj[ i] 是条目的组合除以 l1 和 l2 范数。最后，我需要输出:pj, v, w 用于后续应用。

// X = new double [d*n]; is the input.
double alpha = 0.5;
double *pj = new double[d];
double *x_abs = new double[d];
double *x_2 = new double[d];
double *v = new double[n]();
double *w = new double[n]();
for (unsigned long j=0; j<n; ++j) {
        jm = j*m;
        jd = j*d;
        for (unsigned long i=0; i<d; ++i) {
            x_abs[i] = abs(X[i+jd]);
            v[j] += x_abs[i];
            x_2[i] = x_abs[i]*x_abs[i];
            w[j] += x_2[i];    
        }
        for (unsigned long i=0; i<d; ++i){
            pj[i] = alpha*x_abs[i]/v[j]+(1-alpha)*x_2[i]/w[j];     
        }

   // functionA(pj){ ... ...}  for subsequent applications
} 
// functionB(v, w){ ... ...} for subsequent applications

我的上述算法需要 O(nd) float /时间复杂度，任何人都可以帮助我通过使用 building-functoin 或 C++ 中的新实现来加速它吗？减少O(nd)中的常数值对我也很有帮助。

Answer 1

让我猜猜:因为你有与性能相关的问题，你的 vector 的维度非常大。
如果是这种情况，那么值得考虑“CPU 缓存局部性”——一些有趣的信息in a cppcon14 presentation .
如果数据在 CPU 缓存中不可用，那么 abs -ing 或平方它一旦可用，与 CPU 等待数据的时间相比就相形见绌了。

考虑到这一点，您可能希望尝试以下解决方案(不保证会提高性能 - 编译器可能在优化代码时实际应用这些技术)

for (unsigned long j=0; j<n; ++j) {
        // use pointer arithmetic - at > -O0 the compiler will do it anyway
        double *start=X+j*d, *end=X+(j+1)*d;

        // this part avoid as much as possible the competition
        // on CPU caches between X and v/w.
        // Don't store the norms in v/w as yet, keep them in registers
        double l1norm=0, l2norm=0;
        for(double *src=start; src!=end; src++) {
            double val=*src;
            l1norm+=abs(src);
            l2norm+= src*src;
        }
        double pl1=alpha/l1norm, pl2=(1-alpha)*l2norm;
        for(double *src=start, *dst=pj; src!=end; src++, dst++) {
          // Yes, recomputing abs/sqr may actually save time by not
          // creating competition on CPU caches with x_abs and x_2
          double val=*src;
          *dst = pl1*abs(val) + pl2*val*val;
        }    
        // functionA(pj){ ... ...}  for subsequent applications

        // Think well if you really need v/w. If you really do,
        // at least there are two values to be sent for storage into memory,
        //meanwhile the CPU can actually load the next vector into cache
        v[j]=l1norm; w[j]=l2norm;
}
// functionB(v, w){ ... ...} for subsequent applications