Concatenation of arrays of structure

Hi all

I have to concatenate the field of an array of structure. Here a simple example:

a=struct('a',[]);
a(1).a=[1:5;6:10];
a(2).a=[10:50;60:100];  [EDITED, should be:]  [10:10:50; 60:10:100]

Results:

Concatenated_afield=[1,2,3,4,5,6,7,8,9,10;10,20,30,40,50,60,70,80,90,100]

Thank you

Best regards

[Image Analyst]

You can't do that unless you change the step in (2) to be 10 instead of 1, or change (1) to be a 2-by-41 array like (2) is instead of a 2 by 5 array.

Answer 1

[Image Analyst]

Try this:

% Construct sample data
s=struct('a',[]);
s(1).a=[1:5;6:10]
s(2).a=[10:10:50;60:10:100]
% Now concatenate:
t1 = s(1).a'
t2 = s(2).a'
output = [t1(:), t2(:)]'

In the command window:

t1 =
     1     6
     2     7
     3     8
     4     9
     5    10
t2 =
    10    60
    20    70
    30    80
    40    90
    50   100
output =
     1     2     3     4     5     6     7     8     9    10
    10    20    30    40    50    60    70    80    90   100

Note I didn't use the field name of "a" on a structure also called "a" - I think that's a very bad idea that will lead to confusion, so I named my structure "s".

[pietro]

thanks for your reply. How can I adapt it for a more general solution? My struct array is 500 elements long.

[Image Analyst]

Use a for loop

% Construct sample data
s=struct('a',[]);
s(1).a=[1:5;6:10]
s(2).a=[10:10:50;60:10:100]
s(3).a=[20:10:60;70:10:110]
% Now concatenate
for k = 1 : length(s)
  this_t = s(k).a'
  output(k, :) = this_t(:)';
end
output

Answer 2

[Jan]

a      = struct('a',[]);
a(1).a = [1:5; 6:10];
a(2).a = [10:10:50; 60:10:100];

v = cat(2, a.a);
r = reshape(permute(reshape(v, 2, 5, 2), [3,2,1]), [2, 10]);

Answer 3

[per isakson]

I assume that &nbsp [10:50;60:100] &nbsp should be &nbsp [10:10:50;60:10:100]

    a=struct('a',[]);
    a(1).a=[1:5;6:10];
    a(2).a=[10:10:50;60:10:100];

    >> cat( 1, transpose( a(1).a(:) ), transpose( a(2).a(:) ) )
    ans =
         1     6     2     7     3     8     4     9     5    10
        10    60    20    70    30    80    40    90    50   100

&nbsp

And another try

    transpose(cell2mat(arrayfun(@(s)reshape(transpose(s.a),[],1),a,'uni',false)))
    ans =
         1     2     3     4     5     6     7     8     9    10
        10    20    30    40    50    60    70    80    90   100

And a for-loop

    M = nan( length(a), length(a(1).a(:)) );
    for jj = 1 : length( a)
        M( jj, : ) = [ a(jj).a(1,:), a(jj).a(2,:) ];    
    end

    xlswrite( filespec, M )

[pietro]

Thanks for your help, but the result is different and it works only with arrays with two elements. My array is longer than 500 elements.

[per isakson]

"but the result is different" &nbsp In what way different?

[pietro]

This is the result I need:

      1     2     3     4     5     6     7     8     9    10
      10    20    30    40    50    60    70    80    90   100

[per isakson]

ok - I fixed it

[pietro]

Thanks, but I still get a different result, in the first row I have 1 6 3...instead of 1 2 3.

[per isakson]

Now I think I got it right. It's a bit ridiculous to squeeze it into one line. A plain for-loop is probably better.

[pietro]

Why? I need it for printing the result in one xls file

[per isakson]

Because the for-loop is

• easier to construct
• easier to read and understand in three weeks from now
• and - I guess - executes faster