Convert multi dimensional array to a matrix

Let's say I have A = rand(2,2,2,2). Considering the last two index of the Array A(2,2,x,x) as positions in a matrix such that Matrix(1,1) = A(2,2,1,1) Matrix(1,2) = A(2,2,1,2)..... and so forth.
Ultimately I will end up with a 4,4 matrix in this case.
How can I do this without writing a loop. Please advise.
Update:
I think my question didn't come through correctly. I apologize for it.
Let's have
A(:,:,1,1) =
0.1455 0.8693
0.1361 0.5797
A(:,:,2,1) =
0.5499 0.8530
0.1450 0.6221
A(:,:,1,2) =
0.3510 0.4018
0.5132 0.0760
A(:,:,2,2) =
0.2399 0.1839
0.1233 0.2400
Now I want to have a new Matrix such that
1 2
1 M = [M11 M12;
2 M21 M22];
In my M11 I want A(:,:,1,1)
In my M12 I want A(:,:,1,2) and so forth.
Ultimately I have a 4X4 matrix.
Does it make sense?
Thank you

Réponses (1)

squeeze(A(2, 2, :, :))

6 commentaires

Arif Ahmed
Arif Ahmed le 18 Fév 2017
Sorry. That doesn't work
James Tursa
James Tursa le 18 Fév 2017
Modifié(e) : James Tursa le 18 Fév 2017
Sure it does.
>> A = rand(2,2,2,2)
A(:,:,1,1) =
0.8147 0.1270
0.9058 0.9134
A(:,:,2,1) =
0.6324 0.2785
0.0975 0.5469
A(:,:,1,2) =
0.9575 0.1576
0.9649 0.9706
A(:,:,2,2) =
0.9572 0.8003
0.4854 0.1419
>> squeeze(A(2, 2, :, :))
ans =
0.9134 0.9706
0.5469 0.1419
>> [A(2,2,1,1) A(2,2,1,2); A(2,2,2,1) A(2,2,2,2)] % <-- the explicit elements
ans =
0.9134 0.9706
0.5469 0.1419
You get the same result with the squeeze method and by using the explicit element indexing you specified.
The question is why you think you would end up with a 4x4 matrix. What else about the result are you not telling us? Are you trying to rearrange all of the elements of A somehow? If so, then tell us how.
Arif Ahmed
Arif Ahmed le 18 Fév 2017
Modifié(e) : Arif Ahmed le 18 Fév 2017
I think my question didn't come through correctly. I apologize for it.
Let's have A(:,:,1,1) =
0.1455 0.8693
0.1361 0.5797
A(:,:,2,1) =
0.5499 0.8530
0.1450 0.6221
A(:,:,1,2) =
0.3510 0.4018
0.5132 0.0760
A(:,:,2,2) =
0.2399 0.1839
0.1233 0.2400
Now I want to have a new Matrix such that
1 2
1 M = [M11 M12;
2 M21 M22];
In my M11 I want A(:,:,1,1)
In my M12 I want A(:,:,1,2) and so forth.
Ultimately I have a 4X4 matrix.
Does it make sense?
James Tursa
James Tursa le 18 Fév 2017
Modifié(e) : Stephen23 le 18 Fév 2017
Do you mean just this?
M = [A(:,:,1,1),A(:,:,1,2);
A(:,:,2,1),A(:,:,2,2)];
Your latest example doesn't seem to agree with the indexing for your original post.
Arif Ahmed
Arif Ahmed le 18 Fév 2017
Modifié(e) : Arif Ahmed le 18 Fév 2017
Yes. This is what I meant.
Yes, I agree it doesn't agree with my original post. I was thinking of including 2x2 matrix somehow it translated into (2,2,x,x) in writing. I apologize.
shaziah A
shaziah A le 7 Août 2018
Did you have chance to put them into a matrix? I am also trying to do the same and would really appreciate some help.

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Question posée :

le 17 Fév 2017

Commenté :

le 7 Août 2018

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