How to index every field of a structure and reassign to a structure with a single element in each field

I have a structure where every field is an array of the same length. I need to pass this structure on but only with a single element in each field. I thought of doing it this way

A.b = [1:10];
A.c = [2:11];
A.d = [3:12];
a = structfun(@(x) x(1),A)

but this results in

a = [1;2;3];

The answer I want for the first element is

a.b = 1;
a.c = 2;
a.d = 3;

I will want to run this in a for loop for use in the next function like this

for ii = 1:length(A.b)
     ...
    a = structfun(@(x) x(ii),A); % but modified so that 'a' is a struct like 'A', not an array.
    nextfcn(a);
     ...
end

Answer 1

[KSSV]

A.b = [1:10];
A.c = [2:11];
A.d = [3:12];
a= A ;
for f=fieldnames(a)'
    a.(f{1})(2:end)=[]; 
end

[John Petersen]

Thanks KSSV, Interesting approach. I just had to change the indexing so that I could keep the element I want.

Answer 2

[Chad Greene]

Alternatively, you could stick with the approach you were using, but include the 'uniformoutput',false option.

a = structfun(@(x) x(1),A,'Uni',false)

Answer 3

[George Abrahams]

Old question, but my fieldfun function on File Exchange / GitHub does what you expected structfun to do: output a structure a with the same fields as structure A.

A = struct( 'b', 1:10, 'c', 2:11, 'd', 3:12 );

nthelement = @(n) fieldfun( @(x) x(n), A );
a = nthelement(1)

% a = struct with fields:
%     b: 1
%     c: 2
%     d: 3

a = arrayfun( nthelement, 1:numel(A.b) )

% a = 1×10 struct array with fields:
%     b
%     c
%     d