Assign matrix to struct
63 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
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 commentaires
Rik
le 23 Juil 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
Réponse acceptée
Guillaume
le 23 Juil 2018
Modifié(e) : Guillaume
le 23 Juil 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 commentaires
Jan
le 24 Juil 2018
Modifié(e) : Jan
le 24 Juil 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.
Plus de réponses (1)
KL
le 23 Juil 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
Voir également
Catégories
En savoir plus sur Matrix Indexing dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!