Assign matrix to struct
35 views (last 30 days)
Show older comments
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.
3 Comments
Rik
on 23 Jul 2018
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
Accepted Answer
Guillaume
on 23 Jul 2018
Edited: Guillaume
on 23 Jul 2018
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
4 Comments
Jan
on 24 Jul 2018
Edited: Jan
on 24 Jul 2018
+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.
More Answers (1)
KL
on 23 Jul 2018
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
See Also
Categories
Find more on Structures in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!