Vectorization in more than two dimensions

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Nicole Brimhall
Nicole Brimhall 2020-11-25
评论： Nicole Brimhall 2020-11-25
Is there a way to use vectorized operations in more than two dimensions? For example, in the following code I want to create a three-dimensional matrix based off a function in three variables. I am able to vectorize in two dimensions using the transpose function, but for the third dimension I use a for loop. Is there a way to eliminate the loop?
x=1:1000;x=x';
y=1:586;
z=1:247;
U=zeros(length(x),length(y),length(z));
for cnt=1:length(z)
U(:,:,cnt)=(x.^2+y.^3*z(cnt)).*exp(x*z(cnt).^2).*besselj(0,y);
end
  1 个评论
Nicole Brimhall
Nicole Brimhall 2020-11-25
I think I figured out a solution. If there is a more elegant way, I am all ears. I want to extend this to five dimensions, and keeping track of the permutations is going to be tricky.
x=1:5;
y=1:6;
z=1:3;
X=repmat(x',[1 length(y) length(z)]);
Y=repmat(y,[length(x) 1 length(z)]);
Z=repmat(z,[length(x) 1 length(y)]);Z=permute(Z,[1 3 2]);
U=(X.^2+Y.^3.*Z).*exp(X.*Z.^2).*besselj(0,y);

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回答（1 个）

Walter Roberson
Walter Roberson 2020-11-25
编辑：Walter Roberson 2020-11-25
Ran in:
X = reshape(1:5, [], 1, 1);
Y = reshape(1:6, 1, [], 1);
Z = reshape(1:3, 1, 1, []);
U = (X.^2+Y.^3.*Z).*exp(X.*Z.^2).*besselj(0,Y);
size(U)
ans = 1×3
5 6 3
●
Need R2016b or later.
  2 个评论
Jan
Jan 2020-11-25
This is much more efficient then expanding the arrays. besselj is expensive. Then expanding the argument wastes a lot of time with calculating the results for the same inputs.

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