# Creating different types of arrays with conditions, then make an array with the combination of all elements, and finally separate the arrays with a sequence

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Md. Asadujjaman on 18 Jan 2022
Answered: _ on 18 Jan 2022
Suppose,
x and y are two variables. I want to make x number of arrays each containing y elements where each array elements will be multiplied by (1000*x). Then, I want to make an array (say A) with the combination of the x array elements.
For example, x=3 and y=4, then there will be three arays (say a1, a2, a3) each will contain y elements i.e. [1 2 3 4] multiplied by (1000*x).
i.e., for
x=1, a1=[1001 1002 1003 1004]
x=2, a2=[2001 2002 2003 2004]
x=3, a3=[3001 3002 3003 3004]
Now, the array (A) will be the combination of a1, a2 and a3; for instance
A=[1001 2003 2004 3001 3004 1002 2001 2002 1004 1003 3002 3003 ]
(Q1) How can I construct the array A easily?
Later, I want to sequence a1, a2 and a3 based on the sequence of their elements in A.
i.e.,
a1=[1001 1002 1004 1003]
a2=[2003 2004 2001 2002]
a3=[3001 3004 3002 3003]
(Q2) How can I construct a1, a2, a3 from the array A easily?

_ on 18 Jan 2022
x = 3;
y = 4;
a = 1000*(1:x).'+(1:y)
a = 3×4
1001 1002 1003 1004 2001 2002 2003 2004 3001 3002 3003 3004
A = a(randperm(x*y))
A = 1×12
2003 3003 3001 1002 2001 2002 1001 3004 2004 1003 3002 1004
[~,idx] = ismember(A,a);
[~,new_idx] = sort(mod(idx-1,x)+1);
a_new = reshape(a(idx(new_idx)),[],x).'
a_new = 3×4
1002 1001 1003 1004 2003 2001 2002 2004 3003 3001 3004 3002