Assign matrix to struct

Hi everyone

Can you help me for my code?.

I want to assign matrix b to struct g.a at position c.

g(1).a=[1 2 3 4]'
g(2).a=[1 1 3 4]'
g(3).a=[4 3 1 2]'
c=[1 2]
b=[ 3 3 3 3;4 5 4 4]'

My hope output is g.a =

     3     4     4
     3     5     3
     3     4     1
     3     4     2

Thank you so much.

[Jan]

g.a cannot be a matrix, because g is a struct array already. Then g.a means g(:).a and it is not clear, what you want to acieve.

[Rik]

I agree with Jan. If you want to assign vectors to specific fields, you can always use a for-loop, but how you would get your matrix out is unclear to me.

g(1).a=[1 2 3 4]';
g(2).a=[1 1 3 4]';
g(3).a=[4 3 1 2]';
c=[1 2];
b=[ 3 3 3 3;4 5 4 4]';
for n=1:numel(c)
    g(c(n)).a=b(:,n);
end

[Tiki Tiki]

Thank. but actualy my problem is speed up. so i dont use For here. Can you help me more?

Answer 1

[Guillaume]

As per jan's comment, g.a means nothing. If you type g.a in the command window, you will get 3 outputs (3 times: ans = ...).

[g.a] will indeed return a matrix, which is the horizontal concatenation of all the outputs of g.a. However, [g.a] does not exist in the structure. I believe that you've misunderstood how structures work if that's what you want.

If you want to replace g(c(i)).a by b(:, i), then the easiest is a loop:

for col = 1:numel(c)
   g(c(col)).a = b(:, col);
end

While it can be done without a loop, it's awkward and probably slower:

a = {g.a};  %concatenate all in a cell array
a(c) = num2cell(b, 1);  %replace columns determined by c by columns in b
[g.a] = a{:};   %put back into structure

[Tiki Tiki]

Thank. but my problem is speed up. so when i use num2cell function. it takes much time in my code. so i just want to use vectorztion in this case to speed up.

do you have any solution for this? Thank you.

[Guillaume]

The loop should be the fastest option. You cannot vectorise the loop due to your usage of a structure array.

If you want more speed, then you need to change the way you store your data. It doesn't sound like a structure array is what you need in this case. What KL suggested, a scalar array with matrices in the field would probably work better for you, but since you haven't described your use case, it's hard to tell.

[Jan]

+1. "If you want more speed, then you need to change the way you store your data" - exactly. Speed is not only concerned by the code, but the underlying representation of the data plays an important role.

If g is large, care for a proper pre-allocation. Create e.g. the last field g(max(c)).a at first. If c is sorted, you can run the loop in backward direction:

for col = numel(c):-1:1

But the fastest way would be to keep the matrix b and store only the column indices in the struct array.

[Tiki Tiki]

Thank. Can I ask more about GPU?

How can I work this code on GPU? in my case, g.a is big size. and

because I work on each position of g.a, like get g(1).a or g(2).a,

so How do i change data?

Answer 2

[KL]

probably you mean something like this

g.a(:,1)=[1 2 3 4]';
g.a(:,2)=[1 1 3 4]';
g.a(:,3)=[4 3 1 2]'
c=[1 2]
b=[3 3 3 3;4 5 4 4]'

and then simply use c as indices in a

g.a(:,c) = b

[Tiki Tiki]

Sorry. I dont mean that. because this is a samples in my work. I defined code like that. so i cannot change. Can you help me?