3d matrix initialization and find the specific values after the for loops

1 次查看（过去 30 天）
Saki
Saki 2024-4-20
编辑： Saki 2024-4-24
Hello my Vf is 42444 *1*1000 matrix. I used this code for finding values for the location of 312 and incrementing of 324 from the 42444 values. But why the final matrix A is 131 * 3*1000. I am expecting 131*1*1000.

请先登录，再回答此问题。

采纳的回答

Voss
Voss 2024-4-20
编辑：Voss 2024-4-20

You give A three columns when you do this:

A=[m,n,p]; % size 1x3

Perhaps you meant:

A=zeros([m,n,p]);

But that would give A too many rows.

Anyway, the whole thing can be done in one line:

A = Vf(312:324:end,:,:);
  2 个评论
Saki
Saki 2024-4-20
编辑：Saki 2024-4-24
I got the things what i want.
No need to assign a initial empty A 3d matix.
Voss
Voss 2024-4-20
编辑：Voss 2024-4-20
Ran in:
It's a good idea to clear (if in a script) or preallocate A before your loops.
Or avoid all of that and do:
A = Vf(312:324:end,:,:);
Demonstration:
Loop method
% Vf= is a 3d matix of 42444 * 1 * 1000;
Vf = rand(42444,1,1000);
tic
m = size(Vf, 1);
n = size(Vf, 2);
p = size(Vf, 3);
for j = 1:p
index = 1;
for i = 312:324:m
% Assign the value from Vf to the corresponding location in A
A(index, :, j) = Vf(i, :, j);
index = index + 1;
end
end
toc
Elapsed time is 0.176245 seconds.
Direct method
tic
A_test = Vf(312:324:end,:,:);
toc
Elapsed time is 0.002738 seconds.
The result is the same
isequal(A,A_test)
ans = logical
1
and the "Direct" method is much simpler code and about 65 times faster.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Loops and Conditional Statements 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by