How can I make an array with randomly choose from another matrix?

A=[1 2 7 9] => how to build A like this> A=[1 1 2 2 7 7 9 9]

B=[3 5 6 8 10 11] : randomly choose all items from Band input between 2 same number of A

for example : C=[1 8 6 1 2 5 3 2 7 11 7 9 10 9]

Answer 1

[KALYAN ACHARJYA]

First One:

A=[1 2 7 9] => how to build A like this> A=[1 1 2 2 7 7 9 9]

result=repelem(A,2)

2nd One:

>> A=[1 2 7 9];
>> B=[3 5 6 8 10 11];
>> c1=[A,B];
>> C=c1(randperm(10))
%.................^.....length Required
C =

     5     1     6     9    10     3     8     7    11     2

Or Check here

>> A=[1 2 7 9];
>> A=repelem(A,2)
A =
     1     1     2     2     7     7     9     9
>> B=[3 5 6 8 10 11];
>> c1=[A,B];
>> C=c1(randperm(length(c1)))
C =
     8     1    11     5    10     9     3     1     7     6     9     7     2     2

[Hang Vu]

Thank you so much!

[KALYAN ACHARJYA]

If the question is answered accurately, you can give credit to the answer by accepting it.

[Hang Vu]

Thank you, Do have anyway to put random of B between same number of A? like 1...1 2...2 7...7 9 ....9

[KALYAN ACHARJYA]

In my answer Its random, Random permutation

As per in your example

for example : C=[1 8 6 1 2 5 3 2 7 11 7 9 10 9]

Do have anyway to put random of B between same number of A? like 1...1 2...2 7...7 9 ....9

That may also possible, I have to figure out. Or you can open the new thred for that issue.

[Hang Vu]

we don't have any way to control their position? in my example, B elements are in between the same number

[Hang Vu]

Thank you for your time! what do you mean by" open new thred'? If it helps me to find the solution, could you please tell me how then!

[KALYAN ACHARJYA]

New thred menas new Question.

I have already seen you have posted the new question, with specific pattern.

I will try to answer, if I can generate that C Pattern from v1,v2,v3,v4

As I got the logic of pattern generation of C from v1,v2,v3,v4.

If Time allow I will answer as earliest on that section, or you may get the answer from others members.

Ley see..

[Hang Vu]

Thank you so much sir! This is just my idea, I don't know if it's possible. I will also try to solve it and looking forward for your way!