How I can combine this cell values in a single matrix?

How I can get in a single (12X28) matrix

[Simon Chan]

If the size of the first cell is [5x7 double], you may combine them to become a (12x7) matrix because all of them have 7 columns only.

Or you would like to repeat the cells until the final one has 28 columns?

[Sourasis Chattopadhyay]

First cell is [1X7 double].

I am looking for (12X28) = Row(1*3*2*2) and Column (7+7+7+7).

as example,

Answer 1

[Wan Ji]

If X{1} is 1x7 array, the simplest way is

matrixOut = cell2mat(X')

[Sourasis Chattopadhyay]

Thank you for your answer , I have already tried this syntex but it is not full fill my requirement .

I am looking for this combination (12X28) = Row(1*3*2*2) and Column (7+7+7+7).

[Wan Ji]

RowNum = 12;
A = [];
for i = 1:numel(X)
    Y = X{i};
    A = [A, repmat(Y, RowNum/size(Y,1), 1)];
end
% A is what you want

[Wan Ji]

@Sourasis Chattopadhyay

[Wan Ji]

Or you can do by following command:

RowNum = 12;
X = arrayfun(@(i)repmat(X{i},RowNum/size(X{i},1),1), 1:1:numel(X), 'uniform', 0);
A = cell2mat(X) % A is what you want

[Wan Ji]

If your rowNum is a variable, is it calculated in this way?

RowNum = prod(arrayfun(@(i)size(X{i},1), 1:1:numel(X)))

Then

RowNum =

    12 

The whole code my be

RowNum = prod(arrayfun(@(i)size(X{i},1), 1:1:numel(X)));
X = arrayfun(@(i)repmat(X{i},RowNum/size(X{i},1),1), 1:1:numel(X), 'uniform', 0);
A = cell2mat(X) % A is what you want   

[Wan Ji]

@Sourasis Chattopadhyay

Answer 2

[Simon Chan]

Something like this and final matrix is C

clear; clc;
X = cell(1,4);
X{1} = round(10*rand(1,7));                                 
X{2} = round(10*rand(3,7));
X{3} = round(10*rand(2,7));
X{4} = round(10*rand(2,7));
B{1} = repmat(X{1},12,1);
B{2} = repmat(kron(X{2},ones(2,1)),2,1);
B{3} = repmat(X{3},6,1);
B{4} = kron(X{4},ones(6,1));
C = cell2mat(B);

[Sourasis Chattopadhyay]

If my cell size is cange in this case It is not possiable to do by manually . In case my cell size is (mXn) ,then what should I do?

[Simon Chan]

Need to know the pattern how to repeat the matrix. If a pattern can be found, it is able to implement it in a loop.

However, I am also confused since the repeating method of a 2 rows matrix may be different (The left and right matrix in your elaboration).