Assign matrix to struct
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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 个评论
Jan
2018-7-23
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
2018-7-23
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
2018-7-24
采纳的回答
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 个评论
Tiki Tiki
2018-7-24
Guillaume
2018-7-24
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.
+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
2018-7-25
更多回答（1 个）
KL
2018-7-23
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
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