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Matrices in a matrix.

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Serra Aksoy
Serra Aksoy on 6 Feb 2020
Commented: Serra Aksoy on 7 Feb 2020
Hi,
I have 4 matrices: X1=[1;2;3;4] X2=[ 5,6,7,8] X3=[9,10,11,12] X4=[13,14,15,16]
I would like to compare these matrices to each other and i want to create a new matrix depending on this comparison.
So I want to create a Y=[yij] 4x4 matrix which consists of matrices.
Now i'll explain the comparing rule with an example.
Here min(X1)= 1 and the max(X1)=4
Let's take x ∈ X2.
If x satisfy min(X1) <= x <= max(X1) it should be 1 in a new matrix , if it does not satisfy min(X1)<= x <= max(X1) it should be 0 in a new matrix.
So in this example i need a matrix :[ 0; 0; 0; 0 ]
According to this rule, i want to create a matrix Y=[yij] which consist of these comparison matrices.
So the first row of Y should consist of 4 elements in other words 4 different matrices related to X1. ( comparison of X1 with itself and others )
  • First element of this matrix is y11 which consists of comparing X1 with itself. ( y11= [1;1;1;1], since every element satisfies the rule)
  • Second element is y12 which consists of comparing X1 with X2. (y12=[0;0;0;0], since none of the elements satisfies the rule)
  • Third element is y13 which consists of comparing X1 with X3.
  • Last element of the first row is y14 which consists of comparing X1 with X4.
In this way, i want to create a 4x4 matrix Y.
Is there any way to do this with for loop?
Thanks.

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Accepted Answer

Guillaume
Guillaume on 6 Feb 2020
Edited: Guillaume on 6 Feb 2020
First thing: a matrix of matrices does not exist mathematically, so what you're asking is an impossibility. In matlab, you can store matrices in a cell array. cell arrays are not matrices and work differently so may not be what you want. For example you can't perform mathematical operations on a cell array.
Second thing: Numbered variables are always a bad idea and will greatly complicate your code. You're embedding an index in the variable name which is problematic. Instead you should be using a cell array or a 2D matrix to store your vectors. Either way, you then use standard matlab indexing to refer to each vector.
I would recommend you store your X vectors as rows of a matrix:
X = [1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16];
X(1, :) is your original X1, etc.
With that storage, what you want can be done easily without a loop. As said it can't be stored as a 2D matrix. I'd recommend a 3D matrix instead
%Better X for demo
X = [1 2 3 4 5; ... x1
3 5 7 9 11; ... x2
2 4 6 8 10; ... x3
3 9 12 15 18]; % x4
minx = min(X, [], 2); %min of each vector
maxx = max(X, [], 2); %max of each vector
result = X >= reshape(minx, 1, 1, []) & X <= reshape(maxx, 1, 1, [])
result is a numvector x numcolumn x numvector matrix, where result(r, :, p) is the comparison of X(r, :) with X(p, :).

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Serra Aksoy
Serra Aksoy on 6 Feb 2020
Thank you very much for your quick response.
But i have a question.
X = [1 2 3 4 5
3 5 7 9 11
2 4 6 8 10
3 9 12 15 18];
Here when i compare X1 with itself, for a result i need [1 1 1 1 1] ( Since every element of X1 satisfies min(X1)<= x <=max(X1))
or when i compare X1 with X2, the result should be [ 1 1 0 0 0 ] ( Since only the first two elements of X2, 3 and 5, satisfy the condition: min(X1) <= x <= max (X1) )
So in your code, when i try to compare X1 with X1 using result(1, :, 1), i cant get the result i want.
Guillaume
Guillaume on 6 Feb 2020
After fixing several typos in the code, you'll find it works properly. Sorry, I should have tested it first.
You'll see that result(1, :, 1) (X1 compared with X1) is indeed [1 1 1 1 1] and result(1, :, 2) (X1 compared with X2) is indeed [1 1 0 0 0]
Serra Aksoy
Serra Aksoy on 7 Feb 2020
Thank you very much for your help. But i have one more queistion.
Could you please just take a look at it? Thanks.

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