Replacing Values of a Larger Matrix with that of a Smaller Matrix

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Richard Zareck
Richard Zareck 2017-9-10
回答： Jose Marques 2017-9-10
I have a matrix A that looks like:
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
and a matrix B that looks like:
0 0
0 0
0 0
0 0
0 0
0 0
How can I create a new matrix C that looks like:
0 0 1 1 0 0 1 1 0 0
0 0 1 1 0 0 1 1 0 0
0 0 1 1 0 0 1 1 0 0
0 0 1 1 0 0 1 1 0 0
0 0 1 1 0 0 1 1 0 0
0 0 1 1 0 0 1 1 0 0
by inserting B into A and without using loops. Also if this is not possible, is there a way to create matrix C from matrix A?

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采纳的回答

Cedric
Cedric 2017-9-10
编辑：Cedric 2017-9-10
b = mod( floor( (0:size( A, 2 )-1)/2 ), 2 ) ;
C = bsxfun( @times, A, b ) ;
Note that the way I define b=00110011.. is ... shows that I am tired!
  2 个评论
Cedric
Cedric 2017-9-10
My pleasure. I answered thinking that you wanted to "mask" columns of A, hence the multiplication by 0 and 1 of given columns. Let me know if the purpose was different.

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更多回答（1 个）

Jose Marques
Jose Marques 2017-9-10
imgSiz = [x1,y1];
blkSiz = [x1,size_column];
numRep = imgSiz./blkSiz;
basMat = toeplitz(mod(0:numRep(1)-1,2),mod(0:numRep(2)-1,2));
mask(:,:) = repelem(basMat,x1,size_column);

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