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

2019 3 18
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更新时间：2019 3 18
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0 个投票

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

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Geoff Hayes
Geoff Hayes 2019-3-18
parag - do all 2x2 zero matrices get replaced with the same 2x2 (non-zero) matrix? Or are there different replacement matrices?
parag gupta
parag gupta 2019-3-18
编辑：parag gupta 2019-3-18
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
Guillaume 2019-3-18
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
parag gupta 2019-3-18
编辑：Guillaume 2019-3-18
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
Matt J 2019-3-18
编辑：Matt J 2019-3-18
At end I will get 12 matrices...
No you will not. The code you have shown produces 3 matrices p1,p2,p3.
parag gupta
parag gupta 2019-3-18
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
Guillaume 2019-3-18
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
parag gupta 2019-3-18
编辑：parag gupta 2019-3-18
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 :)

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 采纳的回答

Matt J
Matt J 2019-3-18
编辑：Matt J 2019-3-18

0 个投票

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)) );

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parag gupta
parag gupta 2019-3-18
编辑：parag gupta 2019-3-18
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
Matt J 2019-3-18
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
parag gupta 2019-3-18
thank you so much . :)
haha..sorry for the confusion.
thanks again for all your help..love u :)

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更多回答（1 个）

Matt J
Matt J 2019-3-18
编辑：Matt J 2019-3-18

0 个投票

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;

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parag gupta
parag gupta 2019-3-18
编辑：parag gupta 2019-3-18
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
Matt J 2019-3-18
编辑：Matt J 2019-3-18
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
parag gupta 2019-3-18
yeah, i have 12 different matrices and want to replaces 12 zeros sub matrices of B with those 12 different matrices...
Matt J
Matt J 2019-3-18
编辑：Matt J 2019-3-18
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
parag gupta 2019-3-18
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?
parag gupta
parag gupta 2019-3-18
Unable to perform assignment because the left and right sides have a different number of elements.
got this error. ;(
Matt J
Matt J 2019-3-18
How many elements does the right hande side have?

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