Copy of one field of a structured array

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.

Answer 1

[per isakson]

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
    >>

[John Petersen]

Thanks per! You've made the 1000 milestone.

Answer 2

[the cyclist]

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!

[per isakson]

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.

[John Petersen]

Thanks to the cyclist. This is actually what I was looking for!

Answer 3

[Azzi Abdelmalek]

   C=struct('B',{A.B})