Why is x(:) so much slower than reshape(x,N,1) with complex arrays?

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Matt J
Matt J on 27 Jul 2021
Edited: Matt J on 26 May 2022 at 16:33
The two for loops below differ only in the flattening operation used to obtain A_1D . Why is the run time so much worse with A_3D(:) than with a call to reshape()?
Nx = 256;
Ny = 256;
Nz = 128;
N = Nx*Ny*Nz;
A0 = rand(N,1);
tic
for k = 1:20
B = reshape( A0, [Nz,Ny,Nx] ) ;
A_3D = fftn(B);
A_1D = reshape( A_3D, N,1); %<--- Version 1
end
toc
Elapsed time is 3.770859 seconds.
tic
for k = 1:20
B = reshape( A0, [Nz,Ny,Nx] ) ;
A_3D = fftn(B);
A_1D = A_3D(:); %<--- Version 2
end
toc
Elapsed time is 5.056827 seconds.
  7 Comments
Bruno Luong
Bruno Luong on 28 Jul 2021
I must admit that understanding why/when MATLAB make data copy become obscure to me since few years now. I did not come to a full understanding of how it works.

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Accepted Answer

Matt J
Matt J on 28 Jul 2021
The following simple test seems to support @Bruno Luong's conjecture that (:) results in data copying. The data of B1 resulting from reshape() has the same data pointer location as A, but B2 generated with (:) points to different data.
format debug
A=complex(rand(2),rand(2))
A =
Structure address = 7f3f47f4e0e0 m = 2 n = 2 pr = 7f3fcb0112e0 0.5114 + 0.6181i 0.5881 + 0.4450i 0.5713 + 0.9018i 0.3682 + 0.8103i
B1=reshape(A,4,1),
B1 =
Structure address = 7f3fcf1f4be0 m = 4 n = 1 pr = 7f3fcb0112e0 0.5114 + 0.6181i 0.5713 + 0.9018i 0.5881 + 0.4450i 0.3682 + 0.8103i
B2=A(:)
B2 =
Structure address = 7f3f47e45a20 m = 4 n = 1 pr = 7f3faff0b980 0.5114 + 0.6181i 0.5713 + 0.9018i 0.5881 + 0.4450i 0.3682 + 0.8103i
  8 Comments
Walter Roberson
Walter Roberson on 14 Aug 2021
The (:) options are the slowest. reshape(abs(A),N,1) might possibly be the fastest -- there is notable variation in different runs.
Nx = 256;
Ny = 256;
Nz = 128;
N = Nx*Ny*Nz;
A0 = complex(randn(Nx, Ny, Nz), randn(Nx, Ny, Nz));
t(1) = timeit(@() use_abs_all(A0, N), 0)
t = 0.0937
t(2) = timeit(@() use_abs_colon(A0, N), 0)
t = 1×2
0.0937 0.1727
t(3) = timeit(@() use_abs_reshape_null(A0, N), 0)
t = 1×3
0.0937 0.1727 0.0994
t(4) = timeit(@() use_abs_reshape_N(A0, N), 0)
t = 1×4
0.0937 0.1727 0.0994 0.0935
t(5) = timeit(@() use_all(A0, N), 0)
t = 1×5
0.0937 0.1727 0.0994 0.0935 0.1012
t(6) = timeit(@() use_colon(A0, N), 0)
t = 1×6
0.0937 0.1727 0.0994 0.0935 0.1012 0.1802
t(7) = timeit(@() use_reshape_null(A0, N), 0)
t = 1×7
0.0937 0.1727 0.0994 0.0935 0.1012 0.1802 0.1013
t(8) = timeit(@() use_reshape_N(A0, N), 0)
t = 1×8
0.0937 0.1727 0.0994 0.0935 0.1012 0.1802 0.1013 0.1018
cats = categorical({'abs(all)', 'abs(:)', 'reshape(abs,[])','reshape(abs,N)', 'all', '(:)', 'reshape([])', 'reshape(N)'});
bar(cats, t)
function B = use_abs_all(A, N)
B = max(abs(A), [], 'all');
end
function B = use_abs_colon(A, N)
B = max(abs(A(:)));
end
function B = use_abs_reshape_null(A, N)
B = max(reshape(abs(A), [], 1));
end
function B = use_abs_reshape_N(A, N)
B = max(reshape(abs(A), N, 1));
end
function B = use_all(A, N)
B = max(A, [], 'all');
end
function B = use_colon(A, N)
B = max(A(:));
end
function B = use_reshape_null(A, N)
B = max(reshape(A, [], 1));
end
function B = use_reshape_N(A, N)
B = max(reshape(A, N, 1));
end

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More Answers (2)

Walter Roberson
Walter Roberson on 28 Jul 2021
Nx = 256;
Ny = 256;
Nz = 128;
N = Nx*Ny*Nz;
A0 = rand(Nx, Ny, Nz);
timeit(@() use_colon(A0, N), 0)
ans = 8.3490e-06
timeit(@() use_reshape_null(A0, N), 0)
ans = 6.5490e-06
timeit(@() use_reshape_N(A0, N), 0)
ans = 6.0925e-06
function use_colon(A, N)
B = A(:);
end
function use_reshape_null(A, N)
B = reshape(A, [], 1);
end
function use_reshape_N(A, N)
B = reshape(A, N, 1);
end
In this particular test, the timing is close enough that we can speculate some reasons:
Using an explicit size to reshape to is faster than reshape([]) because reshape([]) has to spend time calculating the size based upon dividing numel() by the size of the known parameters.
Using (:) versus reshape() is not immediately as clear. The model for (:) is that it invokes subsref() with struct('type', {'()'}, 'subs', {':'}) and then subsref() has to invoke reshape() . I point out "model" because potentially the Execution Engine could optimize all of this, and one would tend to think that optimization of (:) should be especially good.
  10 Comments
Adam Danz
Adam Danz on 12 Aug 2021
I noticed that when I re-run it within a script without clearing variables, the second peak at x=5 vanishes. Still curious but out of ideas.

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Matt J
Matt J on 26 May 2022 at 16:29
Edited: Matt J on 26 May 2022 at 16:33
I was just told by Tech Support that the issue was fixed in R2022a, but it doesn't appear that way:
Nx = 256;
Ny = 256;
Nz = 128;
N = Nx*Ny*Nz;
A0 = rand(Nx, Ny, Nz);
A0=complex(A0,A0);
timeit(@() A0(:), 0)
ans = 0.0530
timeit(@() use_reshape_null(A0, N), 0)
ans = 6.5199e-06
timeit(@() use_reshape_N(A0, N), 0)
ans = 6.8033e-06
function use_reshape_null(A, N)
B = reshape(A, [], 1);
end
function use_reshape_N(A, N)
B = reshape(A, N, 1);
end

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