How to replace 2x2 zero sub matrices by different 2x2 matrices ?

How to replace 2x2 zero submatrix of B with 2x2 matrices different matrice.I have 12 matrices of size 2x2 and want to replace zeros of B by those matrices.

B =

0.8776 0.6513 0 0 0 0 0 0

0.0144 0.8646 0 0 0 0 0 0

0 0 0.2943 0.0560 0 0 0 0

0 0 0.1799 0.8169 0 0 0 0

0 0 0 0 0.9263 0.5289 0 0

0 0 0 0 0.0682 0.6944 0 0

0 0 0 0 0 0 0.5811 0.2124

0 0 0 0 0 0 0.6372 0.5433

Thanks

[Geoff Hayes]

parag - do all 2x2 zero matrices get replaced with the same 2x2 (non-zero) matrix? Or are there different replacement matrices?

[parag gupta]

yeah,with the same 2x2 (non-zero) matrix.

for eg i have p1 (2x2 matrix)

p1 = a b

c d

B =

0.8776 0.6513 0 0 0 0 0 0

0.0144 0.8646 0 0 0 0 0 0

then i want

0.8776 0.6513 a b 0 0 0 0

0.0144 0.8646 c d 0 0 0 0

[Guillaume]

Can you describe using matlab syntax (so we can easily copy/paste into matlab) what your inputs are. In your example, above

%this is valid matlab:
B = [0.8776    0.6513         0         0         0         0         0         0
     0.0144    0.8646         0         0         0         0         0         0
          0         0    0.2943    0.0560         0         0         0         0
          0         0    0.1799    0.8169         0         0         0         0
          0         0         0         0    0.9263    0.5289         0         0
          0         0         0         0    0.0682    0.6944         0         0
          0         0         0         0         0         0    0.5811    0.2124
          0         0         0         0         0         0    0.6372    0.5433]

In particular, it's not clear how you define your replacements. Have you got 12 replacements matrices? How are they stored? Easiest would be a cell array:

replacements = {[1 2; 3 4], [5 6; 7 8], etc...}  %not valid syntax currently!

If it's a cell array how do we know where replacement{5} (for example) goes ?

[parag gupta]

N = 8
n = 2
Acell=mat2cell(rand(N,n), ones(1,4)*n,n)
celldisp(Acell)
B=blkdiag(Acell{:})
a = 0.01;
for i = 1:4
    p1= a*Acell{i} 
    p2 =a^2*Acell{i}
    p3 = a^3*Acell{i}
end

At end I will get 12 matrices and want to replace zero sub matrices of B with those 12 matrices.

basically i want solution in this form

x p1 p2 p3

p1 x p2 p3

p1 p2 x p3

p1 p2 p3 x

[Matt J]

At end I will get 12 matrices...

No you will not. The code you have shown produces 3 matrices p1,p2,p3.

[parag gupta]

for i = 1:4

p1= a*Acell{i}

p2 =a^2*Acell{i}

p3 = a^3*Acell{i}

end

where Acell{i} is a 2x2 matrix and a = 0.01

according to this command i will get 3 (p1 matrices),3 (p2 matrices),

3 (p3 matrices),...so at end 12 matrices.

Could u pls cehck this ( i hve removed semi colons)

[Guillaume]

The loop would imply that you want to create 12 matrices. However, since it overwrites the matrices in the previous step, you'll end up with only 3 matrices. The ones created on the last step of the loop.

While you say that you have 12 matrices, your I want solution in this form implies that there are only 3 matrices, which are rotated on each row.

A bit confusing as to what you have.

I doubt it will be difficult to do what you want, but the method will vary depending on exactly what it is that you want. So, please clarify.

[parag gupta]

oops, u r right it will overwites the matrices.Could u pls tell what to do so that I end up with all 12 matrices?

sorry for the confusion, I want this matrix at the end.

x matrix 1 matrix 2 matrix 3

matrix 4 x matrix 5 matrix 6

matrix 7 matrix 8 x matrix 9

matrix 10 matrix 11 matrix 12 x

thanks you so much :)

Answer 1

[Matt J]

basically i want solution in this form

x p1 p2 p3

p1 x p2 p3

p1 p2 x p3

p1 p2 p3 x

C={zeros(2), p1,p2,p3};

result=B+cell2mat( C(toeplitz(1:4))  );

[parag gupta]

Thanks MJ

sorry for the confusion, I want this matrix at the end.

x matrix 1 matrix 2 matrix 3

matrix 4 x matrix 5 matrix 6

matrix 7 matrix 8 x matrix 9

matrix 10 matrix 11 matrix 12 x

[Matt J]

Very similar to things I've already mentioned:

B(B==0)=cat(3,matrix4,matrix7,matrix10,matrix1,matrix8,matrix11,...
              matrix2,matrix5,matrix12,matrix3,matrix6,matrix9);

[parag gupta]

thank you so much . :)

haha..sorry for the confusion.

thanks again for all your help..love u :)

Answer 2

[Matt J]

Here is a simple example, but the right hand side can be any array with an appropriate number of elements

B(B==0)=1:48;

[parag gupta]

for eg i have p1 (2x2 matrix)

p1 = a b

c d

p2 = a1 b1

c1 d1

B =

0.8776 0.6513 0 0 0 0 0 0

0.0144 0.8646 0 0 0 0 0 0

then i want

0.8776 0.6513 a b a1 b1 0 0

0.0144 0.8646 c d a2 b2 0 0

thanks MJ :)

[Matt J]

You said you have 12 matrices in your original post.

B(B==0)=cat(3,p1,p2,p3,...,p12);

Note that it is probably going to continue to be awkward if you hold the matrices in separate variables.

[parag gupta]

yeah, i have 12 different matrices and want to replaces 12 zeros sub matrices of B with those 12 different matrices...

[Matt J]

B(B==0)=cat(3,p1,p2,p3,...,p12);

Note that it is probably going to continue to be awkward if you hold your matrices in separate variables.

[parag gupta]

N = 8
n = 2
Acell=mat2cell(rand(N,n), ones(1,4)*n,n)
celldisp(Acell)
B=blkdiag(Acell{:})
a = 0.01;
for i = 1:4
p1= a*Acell{i} 
p2 =a^2*Acell{i}
p3 = a^3*Acell{i}
end

At end I will get 12 matrices and want to replace zeros of B with that.Is there another way to do this?

[Matt J]

I have showed you how.

[Matt J]

B(B==0)=cat(3,p1,p2,p3,...,p12);

[parag gupta]

Unable to perform assignment because the left and right sides have a different number of elements.

got this error. ;(

[Matt J]

How many elements does the right hande side have?