Assignment between structures with common fields

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Akshay
Akshay 2014-6-27
编辑： pietro 2014-6-30
Hi,
I have two structures A and B:
A = struct ('field1', {1,2}, 'field2', {3,4}, 'field3', {5,6});
B = struct ('field1', {7,8}, 'field2', {9,10});
Therefore A and B are:
A =
1x2 struct array with fields:
field1
field2
field3
B =
1x2 struct array with fields:
field1
field2
I want to assign the values of field1 and field2 from B to corresponding field1 and field2 of A. Structure arrays A and B have the same number of elements.
Is there any way for me to do this with the most efficiency?
Thanks

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回答（2 个）

Sara
Sara 2014-6-27
If you do not want to assign all the fields of A to B, I'd do:
f = {'field1','field2'}; % list of fields
for i = 1:numel(f)
B.(f{i}) = A.(f{i});
end
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Image Analyst
Image Analyst 2014-6-28
Yeah, you'd think. You'd even think the simpler
A.field1 = B.field1
A.field2 = B.field2
would work, but none of them do. Just try it and see.
Akshay
Akshay 2014-6-28
This is not quite what I was looking for, but a variant of this answer gives me a decent answer.
I want to assign values FROM B to A.
I can do the following:
A = struct ('field1', {1,2}, 'field2', {3,4}, 'field3', {5,6});
B = struct ('field1', {7,8}, 'field2', {9,10});
fields = fieldnames(B);
for ctr = 1:numel(fields)
[A.(fields{ctr})] = B.(fields{ctr});
end
Note the [] around the left hand side of the assignment. This was the only change from your answer (apart from assignment order).
Thanks for the help. I'd ideally want something more compact than this, maybe something without the need for a loop.

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Image Analyst
Image Analyst 2014-6-28
Try it this way:
A = struct ('field1', {1,2}, 'field2', {3,4}, 'field3', {5,6})
B = struct ('field1', {7,8}, 'field2', {9,10})
numberOfStructuresA = length(A)
numberOfStructuresB = length(B)
if numberOfStructuresA ~= numberOfStructuresB
message = sprintf('You cannot assign all of the fields because the number of elements is different');
uiwait(warndlg(message));
return;
end
for k = 1 : numberOfStructuresA
A(k).field1 = B(k).field1
A(k).field2 = B(k).field2
end

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