在 octave 上矢量化多维数组填充答案

【问题标题】：Vectorize multi-dimensional array population on octave在 octave 上矢量化多维数组填充
【发布时间】：2019-05-31 02:13:47
【问题描述】：

这是我的实际代码：

Lr = linspace(100,300,10);
vi = linspace(10,30,10);
vf = linspace(10,30,10);
b = linspace(0.5,1.2,10);
h = linspace(0.3,0.8,10);
Rc = [1000, 1500, 2000, 2500, 3000, 3500;
      29, 22.4, 17.3, 13.4, 10.4, 8];
rti = randi(5, 10, 1, 10, 10, 10, 10);
for kk = 1:rows(Lr)
  for jj = 1:length(vi)
    for ll = 1:length(vf)
      for mm = 1:length(b)
        for nn = 1:length(h)
          ratt(kk,1,jj,ll,mm,nn) = Rc(2,rti(kk,1,jj,ll,mm,nn));
        endfor
      endfor
    endfor
  endfor
endfor

在这个例子中，所有向量的长度都是 10，但在实际代码中，它们可以是 1 到 100。如何删除所有这些循环？

【问题讨论】：

标签： multidimensional-array vectorization octave n-dimensional

【解决方案1】：

将所有值作为一维数组从Rc 获取到ratt。
将ratt 重塑为原始代码中的 6D 数组。

代码：

ratt = Rc(2, rti(1:rows(Lr), 1, 1:length(vi), 1:length(vf), 1:length(b), 1:length(h)));
ratt = reshape(ratt, [rows(Lr), 1, length(vi), length(vf), length(b), length(h)]);

测试：为了测试向量化和非向量化代码的相等性和速度，我编写了以下代码：

Lr = linspace(100,300,10);
vi = linspace(10,30,10);
vf = linspace(10,30,10);
b = linspace(0.5,1.2,10);
h = linspace(0.3,0.8,10);
Rc = [1000, 1500, 2000, 2500, 3000, 3500;
      29, 22.4, 17.3, 13.4, 10.4, 8];
rti = randi(5, 10, 1, 10, 10, 10, 10);

disp('Non-vectorized:')
tic
for kk = 1:rows(Lr)
  for jj = 1:length(vi)
    for ll = 1:length(vf)
      for mm = 1:length(b)
        for nn = 1:length(h)
          ratt(kk,1,jj,ll,mm,nn) = Rc(2,rti(kk,1,jj,ll,mm,nn));
        endfor
      endfor
    endfor
  endfor
endfor
toc

disp('')

disp('Vectorized:')
tic
ratt1 = Rc(2, rti(1:rows(Lr), 1, 1:length(vi), 1:length(vf), 1:length(b), 1:length(h)));
ratt1 = reshape(ratt1, [rows(Lr), 1, length(vi), length(vf), length(b), length(h)]);
toc

disp('')

if(isequal(ratt,ratt1))
  disp('ratt is equal to ratt1!')
else
  disp('ratt is not equal to ratt1!')
endif

输出：

Non-vectorized:
Elapsed time is 0.16697 seconds.

Vectorized:
Elapsed time is 0.000309944 seconds.

ratt is equal to ratt1!

【讨论】：