Copy of one field of a structured array
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Is there a way to make a copy of one field of a structured array without doing a loop? I have tried but it just results in the first element.
Creation of sample structure
for i=1:5
A(i).A = i;
A(i).B = i+10;
end
>>C = A.B
C =
11
Likewise
>> C = A(1:3).B
C =
11
instead of C(1).B = 11, C(2).B = 12, and C(3).B = 13 like I want.
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Risposta accettata
per isakson
il 12 Mar 2013
Modificato: per isakson
il 12 Mar 2013
One way to do it:
>> C = cell2struct( num2cell([A.B]), {'B'} )
C =
5x1 struct array with fields:
B
>> C(1).B
ans =
11
>> C(2).B
ans =
12
another
>> C = struct( 'B', num2cell([A.B]) )
C =
1x5 struct array with fields:
B
>> C(1).B
ans =
11
>> C(2).B
ans =
12
>>
Più risposte (2)
the cyclist
il 12 Mar 2013
Modificato: the cyclist
il 12 Mar 2013
Yet another way:
[C(1:numel(A)).B] = A.B;
Note that this method will not overwrite other fields of C, if they exist. Some of the methods in other answers will, so be cautious!
2 Commenti
per isakson
il 12 Mar 2013
Modificato: per isakson
il 12 Mar 2013
The two functions, struct and cell2struct, create a new struct and overwrite an existing C.
I'll try to remember that numel(A) in the rhs-expression creates a new struct if C does not exist. And I'll reread the entry on deal in the documentation.
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