Create a matrix on the basis of other matrix

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luca
luca 2019-10-14
评论： Fabio Freschi 2019-10-15
Hi given a vector
SPI= [2 3 4 8 11 13 14 15 16 19 20];
and the array
AA= [1 2 3 4 0 11 15;
0 0 0 8 13 16 0;
0 0 0 0 0 18 0;
0 0 0 0 0 19 0;
0 0 0 0 0 20 0];
I want to create a matrix DD that is AA but with just the element inside SPI. So:
DD = [ 2 3 4 0 11 15;
0 0 8 13 16 0;
0 0 0 0 0 0;
0 0 0 0 19 0;
0 0 0 0 20 0];
may someone help me?
  2 个评论
Walter Roberson
Walter Roberson 2019-10-14
I notice that you delete the first column of AA as it is left without any elements from SPI. Why are you not also deleting the all-zero row 3 ? Or if the rule is that the first row has to contain an element from SPI, why are you not deleting the column [0; 13; 0; 0; 0] ? What are the rules?
luca
luca 2019-10-14
编辑：luca 2019-10-14
the array SPI could vary and contain all the elements of AA, that instead is fixed. So depending on the value inside SPI I want to derive DD.
Sorry for the third column, was my mistake! now it's fixed

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Fabio Freschi
Fabio Freschi 2019-10-14
编辑：Fabio Freschi 2019-10-14
% your data
SPI = [2 3 4 8 11 13 14 15 16 18 19 20];
AA = [1 2 3 4 0 11 14 15;
0 0 0 8 13 16 0 0;
0 0 0 0 0 0 0 0;
0 0 0 0 0 19 0 0;
0 0 0 0 0 20 0 0];
% mask
iMask = ismember(AA,SPI);
% matrix DD with zeros
DD = AA.*iMask;
% now you can filter out the rows and cols of all zeros
DD = DD(:,any(iMask,1)); % cols filter
DD = DD(any(iMask,2),:); % rows filter
  6 个评论
显示 4更早的评论
Walter Roberson
Walter Roberson 2019-10-15
Fabio's suggested code will eliminate the row 0 0 0 0 0 0;
I do not understand at the moment why that row is being kept in the desired solution.
Fabio Freschi
Fabio Freschi 2019-10-15
Me neither! but I kept all steps separated to that the OP can pick the parts of his choice

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