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

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Saki
Saki 2024 年 4 月 20 日
編集済み: Saki 2024 年 4 月 24 日
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.

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Voss
Voss 2024 年 4 月 20 日
編集済み: Voss 2024 年 4 月 20 日

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 件のコメント
Saki
Saki 2024 年 4 月 20 日
編集済み: Saki 2024 年 4 月 24 日
I got the things what i want.
No need to assign a initial empty A 3d matix.
Voss
Voss 2024 年 4 月 20 日
編集済み: Voss 2024 年 4 月 20 日
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.

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