multiply each element of structured array, assignment of structured array
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I'm not exactly sure if I'm using the right terminology. I have
A(1).B = 5;
A(2).B = 6;
A(3).B = 7;
>>A.B
ans =
5
ans =
6
ans =
7
I have two questions: 1. I want to multiply each element of the structured array so that A.B*2 would equal 10,12,14 instead of 5,6,7. but instead I get the error
>>A.B*2
Error using *
Too many input arguments.
2. How do I make a copy of the structured array? I have tried but it just results in the first element.
>>C = A.B
C =
5
Likewise
>> C = A(1:3).B
C =
5
instead of C(1).B = 5, C(2).B = 6, and C(3).B = 7 like I want.
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Accepted Answer
per isakson
on 11 Mar 2013
Edited: per isakson
on 11 Mar 2013
The short answer is NO, it is not possible. However, ...
>> [A.B] = split( 2*[A.B] )
A =
1x3 struct array with fields:
B
>> A.B
ans =
10
ans =
12
ans =
14
where
function varargout = split( vec )
assert( nargout == numel( vec ) ...
, 'split:narginNargoutMismatch' ...
, 'The number of outputs should match the length of the input vector.')
varargout = num2cell( vec );
end
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More Answers (1)
Jan
on 11 Mar 2013
Edited: Jan
on 11 Mar 2013
C = [A(1:3).B];
% or C = [A(:).B]
% or C = [A.B]
This replies [5,6,7]. To get C(1).B = 5, C(2).B = 6, C(3).B = 7:
C = A;
A direct multiplication is not possible. Structs are simply not designed for arithmetic operations, because this can be done with numerical arrays much better. But it is possible with a loop:
F = fieldnames(A);
for iF = 1:length(F)
aF = F{iF};
A.(aF) = A.(aF) * 2;
end
2 Comments
Ella
on 7 Jun 2023
Hi! So I have a structure with a variable amount of entries. It will always be greater than 1. These stats (Area, MinFeret.. etc) need to be converted to mm. I tried to use the for loop approach, but it says there are too many inputs. Here is my code.
F= fieldnames(stats);
for iF = 1:length(F)
aF = F{iF};
stats.(aF) = stats.(aF) .* pixelSize;
end
Any advice for how I can make it work or another way to do this?
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