How to sample single values from field in *non-scalar structure array*?

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Manuel A. Diaz
Manuel A. Diaz 2023-7-25
评论： Manuel A. Diaz 2023-7-26
Hello Matlab experts,
I'm working on the optimization of some code following the Profiler's measurements.
Here, one main bottleneck is the sampling of an individual element on each of the fields contained within the struct array.
For context, let's consider that a non-scalar struct array with uniform mat sparse fields is built as
A = repmat(struct('mat',speye(3)),10,1);
for k = 1:10
A(k).mat = A(k).mat * k;
end
For the sake of simplicity, I'm interested in sampling each field at indexes (i,j) = (2,2) as
col = zeros(10,1);
for k = 1:10
col(k) = A(k).mat(2,2);
end
So that for the next step, I report each sampled point into a column vector.
Thus the expected output is:
>> col
col =
1
2
3
4
5
6
7
8
9
10
Question: is there a much better (hopefully faster) way to do this? (mainly getting rid of the for-loop)
Disclaimer: I have tried converting this struct array into flat array, [A.mat], or into an array of cells, {A.mat}, but I run into the trouble of the sparse nature of the fields (Get warnings with SPFUN). I'm sure there must be a clever way to do this with arrayfun() or cellfun() but the solution eludes me.
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Manuel A. Diaz
Manuel A. Diaz 2023-7-26
Yes! I'm aware of it.
Although it allowed me to have three indexes, it didn't do the trick on my case ...
Nevertheless thx for the suggestion! : )
-M

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采纳的回答

Manuel A. Diaz
Manuel A. Diaz 2023-7-26
For the sake of documentation, I'm going to respond to my own question.
Turns out (like always) that the solution is very simple:
This problem arises because we are trying to use structures (A(k).mat(i,j)) or cells (A{k}(i,j)) to handle sparse fields like if they werer 3-dimensinal arrays (i.e. A(i,j,k) ). As @Stephen23, commented trying to vectorize this is not currently posible in Matlab. However, the answer for @Bruno Luong showed me that the problem can be simply flatten so that simple indexing can be used.
TL;DR: a simple way to speed up this problem is:
% Get Data
A = repmat(struct('mat',speye(3)),10,1);
for k = 1:10
A(k).mat = A(k).mat * k;
end
% Get Data sizes
[m,n] = size(A(1).mat); % because we have uniformly sized fields
% Compute
A_flat = [A.mat]; % Flatten the structure fields
col = transpose(A_flat(2,2 + 0:m:m*10));
col =
(1,1) 1
(2,1) 2
(3,1) 3
(4,1) 4
(5,1) 5
(6,1) 6
(7,1) 7
(8,1) 8
(9,1) 9
(10,1) 10
Voila !
To probe to you that above idea is correct, I share the screencaps of the profiler measurementes using the original and *this-> solution:
Original Problem:
Using *this-> reformulation
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Bruno Luong
Bruno Luong 2023-7-26
编辑：Bruno Luong 2023-7-26
Ran in:
The speed up depends on the size of mat. Less than 100, the speed up is real; beyond that the flatten runtime increases with the size where-as the for-loop time is constant, independent of the size of the matrix, as showed,in this test script.
matsize = 2.^(1:12);
ntest = length(matsize);
t1 = zeros(1, ntest);
t2 = zeros(1, ntest);
t3 = zeros(1, ntest);
for i = 1:ntest
A = repmat(struct('mat',speye(matsize(i))),10000,1);
for k = 1:numel(A)
A(k).mat = A(k).mat * k;
end
t1(i) = timeit(@() forloop(A), 1);
t2(i) = timeit(@() flatA(A), 1);
t3(i) = timeit(@() afun(A), 1);
end
close all
semilogx(matsize, t1)
hold on
semilogx(matsize, t2)
semilogx(matsize, t3)
xlabel('sizemat')
ylabel('time [s]')
legend('for loop', 'flat mat', 'arrayfun')
function col = forloop(A)
n = numel(A);
col = zeros(n,1);
for k = 1:n
col(k) = A(k).mat(2,2);
end
end
function col = afun(A)
col = arrayfun(@(s) full(s.mat(2,2)), A);
end
function col = flatA(A)
n = numel(A);
% Get Data sizes
m = size(A(1).mat,1); % because we have uniformly sized fields
% Compute
A_flat = [A.mat]; % Flatten the structure fields
col = transpose(A_flat(2,2 + 0:m:m*n));
end
Manuel A. Diaz
Manuel A. Diaz 2023-7-26
Wow ... you went the extra mile too!
Disclosure: I'm working on the range of matsize : [20,400] with my struct size : ~[20^3,400^3].
I have being observing the same this afternoon.
Not safistied with the loop solution. Thus, I'm currently coding mex-file instead ; )
Thx again
-M

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更多回答（1 个）

Bruno Luong
Bruno Luong 2023-7-25
编辑：Bruno Luong 2023-7-25
Ran in:
Just shorter code, not necessary better.
A = repmat(struct('mat',speye(3)),10,1);
for k = 1:10
A(k).mat = A(k).mat * k;
end
col = arrayfun(@(s) full(s.mat(2,2)), A)
col = 10×1
1 2 3 4 5 6 7 8 9 10
  1 个评论
Manuel A. Diaz
Manuel A. Diaz 2023-7-25
Indeed, it works!
... but the profiler indicates that the arrayfun() does little to improve over the for-loop solution.
Nevertheless, I was not aware that full() can be used in this manner. Thus it gives me a hint to reformulate my problem.
Thx much!

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