# Fastest way to differentiate only the sides of large matricies

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Nathan Zechar 2021년 10월 2일
댓글: Nathan Zechar 2021년 10월 4일
I believe I have found one of the fastest ways to differentiate different portions of a large matrix using built in Matlab functions. However, there is a problem to my method, there is a remaining middle value between the side values. Allow me to demonstate with some simple code which makes use of gpuArray. For my case, I need the data on a gpu.
N = 300; % Number of points in x,y,z dimensions
S = 10; % Number of points from side of a dimension
for i = 1:20
B = gpuArray(rand(N,N,N)); % Load some random data
An = diff(B(1:S+1,:,:),1,1); % Differentation from left side
Ap = diff(B(end-S:end,:,:),1,1); % Differentation from right side
A1 = B([2:S+1,N-S+1:N],:,:)-B([1:S,N-S:N-1],:,:);
% The above is the finite difference which is applied to
% both sides of the matrix at the same time. This method
% however, is slower than using built in diff
A2 = diff(B([1:S+1,N-S:N],:,:),1,1);
% The above method is the fastest method, however there is one
% additional data point between S+1 and N - S. Deleting this
% point would add extra computational time
end
However diff is coded on the backend of Matlab, and would need to be modified for my application. If this is true, how would I go about learning how to code up my own diff function for this special case?
Thank you

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### 채택된 답변

Matt J 2021년 10월 3일
편집: Matt J 2021년 10월 3일
This seems to be about as fast, at least on the GTX 1080 Ti.
T=reshape(B([1:S+1,N-S:N],:) ,S+1,2*N^2 );
A3 = reshape( diff(T,1,1), 2*S,N,N);
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Matt J 2021년 10월 4일
편집: Matt J 2021년 10월 4일
No, you don't need to permute.
% along 1st axis
T = reshape(B([1:S+1,N-S:N],:) ,S+1,2,N,N );
A1 = reshape(diff(T,1,1), 2*S,N,N);
% along 2nd axis
T=reshape(B(:,[1:S+1,N-S:N],:),N,S+1,2,N);
A2=reshape( diff(T,1,2) , N,2*S,N);
% along 3rd axis
T=reshape(B(:,:,[1:S+1,N-S:N]),N,N,S+1,2);
A3=reshape( diff(T,1,3) , N,N,2*S);
Nathan Zechar 2021년 10월 4일
@Matt J, thank you again. I was hung up on what was happening on the inside of the first line in the original method.
B([1:S+1,N-S:N],:)
Reduces to 2 dimensions. I thought that was part of the magic and had to permute to that form to get everything to work. Thank you so much again.

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