filling an empty cell dynamically

Question

Lets suppose we have a structure v which has variable number of elements in its objects

>> v(1).a = 1:10;
>> v(2).a = 1:20;
>> v(3).a = 1:30;

Now if we want to put it into a cell array we can simply concatenate and we will have a a with 3 cells with in

>> c = {v(1).a,v(2).a,v(3).a}

c = 

    [1x10 double]    [1x20 double]    [1x30 double]

and we can access any element within cell by using

c{i}(j)

but now for example if i have to fill the same array dynamically in a for loop

c = v(1).a
for i= 2:numel(v)
    c= {c ,v(i).a};
end

for the first 2 iterations it works in the same way as c = {v(1).a,v(2).a}

c = 

[1x10 double]    [1x20 double]

but after the 3rd iteration it converts the first two array into a cell with a cell

c = 

{1x2 cell}    [1x30 double]

how can avoid this ? and create a cell of array instead, like the first case, using a for loop

Answer 1

Try creating c as a cell array and then appending cell arrays to it:

c = {v(1).a}
for i = 2:numel(v)
    c = [c, {v(i).a}];
end

or with a simpler initialisation:

c = {};
for i = 1:numel(v)
    c = [c , {v(i).a}];
end

However, you can do it much more simply with a one-liner:

c = { v.a }

This last solution is a MATLAB "comma-separated list" and is a convenient way to build arrays and cell arrays. If you type v.a at the command line you'll see you get three answers returned (namely v(1).a, v(2).a, v(3).a). By writing { v.a } it's equivalent to writing those three answers, separated by commas, i.e. equivalent to {v(1).a, v(2).a, v(3).a}.

Answer 2

I would recommend this approach:

c = {};
for i = 1:numel(v)
    c{end+1} = v(i).a;
end

or

c = {};
for i = 1:numel(v)
    c(end+1) = {v(i).a};
end

which I suspect (based on this answer) will be faster than the concatenation approach proposed by Justin.