# Problem with cell array appending

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Jaya il 16 Set 2021
Commentato: Stephen23 il 17 Set 2021
mycell is appended with cell arrays in three different areas of my code. Like below
mycell= { }
mycell= A(:,:,1) %1st time. A(:,:,1) is a 1*5 cell array
mycell= {mycell ; B(:,:,1) } %2nd time. B(:,:,1) is a 1*5 cell array
mycell= {mycell ; C(:,:,1) } %3rd time. C(:,:,1) is a 1*5 cell array
1st time output is OK: mycell is a cellarray of 1*5.
2nd time output is also OK: mycell is a 2*1 cell array with each element of 1*5 size.
BUT 3rd time output: mycell is still a 2*1 cell array as below. Why? Why do the previous two elements form as a single element in this third time? Can someone tell me how do I avoid this?
%the output I get after 3rd time line
mycell =
2×1 cell array
{2×1 cell}
{1×5 cell}
% but the output I want is something like.
{1×5 cell}
{1×5 cell}
{1×5 cell}
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Stephen23 il 17 Set 2021
Note the difference:
• {} curly braces creates a cell array, where the inputs are nested inside the new cell array.
• [] square brackets are a concatenation operator. These are used to concatenate any array type.
So if you want to nest cell arrays inside other cell arrays, then use curly braces. But if you want to concatenate any arrays together, use square brackets (or the operators CAT, HORZCAT, VERTCAT).

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### Risposta accettata

Star Strider il 16 Set 2021
Assigning is likely a more efficient approach than concatenation —
A(:,:,1) = randn(1,5);
B(:,:,1) = randn(1,5);
C(:,:,1) = randn(1,5);
mycell{1,:}= A(:,:,1) %1st time. A(:,:,1) is a 1*5 cell array
mycell = 1×1 cell array
{[0.4460 0.5237 0.2281 0.4874 -1.1267]}
mycell{2,:}= B(:,:,1) %2nd time. B(:,:,1) is a 1*5 cell array
mycell = 2×1 cell array
{[ 0.4460 0.5237 0.2281 0.4874 -1.1267]} {[-1.9246 1.0601 0.0070 -0.5618 -0.8078]}
mycell{3,:}= C(:,:,1) %3rd time. C(:,:,1) is a 1*5 cell array
mycell = 3×1 cell array
{[ 0.4460 0.5237 0.2281 0.4874 -1.1267]} {[-1.9246 1.0601 0.0070 -0.5618 -0.8078]} {[ 0.7868 0.1859 0.6091 1.0478 -0.2925]}
This also allows for preallocation, that can significantly improve code efficiency.
The cell concatenation approach creates ‘cells-of-cells’, making the interpretation more difficult. The MATLAB concatenation operator are the square brackets [] so using them will produce the correct result —
mycell2 = { }
mycell2 = 0×0 empty cell array
mycell2 = {A(:,:,1)} %1st time. A(:,:,1) is a 1*5 cell array
mycell2 = 1×1 cell array
{[0.4460 0.5237 0.2281 0.4874 -1.1267]}
mycell2 = [mycell2 ; {B(:,:,1)} ] %2nd time. B(:,:,1) is a 1*5 cell array
mycell2 = 2×1 cell array
{[ 0.4460 0.5237 0.2281 0.4874 -1.1267]} {[-1.9246 1.0601 0.0070 -0.5618 -0.8078]}
mycell2 = [mycell2 ; {C(:,:,1)} ] %3rd time. C(:,:,1) is a 1*5 cell array
mycell2 = 3×1 cell array
{[ 0.4460 0.5237 0.2281 0.4874 -1.1267]} {[-1.9246 1.0601 0.0070 -0.5618 -0.8078]} {[ 0.7868 0.1859 0.6091 1.0478 -0.2925]}
This is less efficient than the indexing approach, because it precludes preallocation.
Experiment to get different results.
.
##### 2 CommentiMostra NessunoNascondi Nessuno
Jaya il 16 Set 2021
Thanks a lot. This solution has made my code interpretation so much easier.
Star Strider il 16 Set 2021
As always, my pleasure!
.

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