Left shift is not working

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Ayesa 2011-12-14

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https://ww2.mathworks.cn/matlabcentral/answers/23974-left-shift-is-not-working

Hi. I am trying to left shift of a m by n matrix but its not working.

   PP = [1 2 3 4           %PP can be any size
         1 2 3 4, 
         1 2 3 4]

I want:

    PP = [1 2 3 4
          2 3 4 0
          3 4 0 0]

I used following code(Here i have to use loop because PP actually a very large size matrrix):

[rows cols]= size(PP);  
B = zeros(size(PP));
for i = 1:rows     
    B = zeros(size(PP));
    PP_shift(i, :) =  [PP(i,(cols:-1:i)) B(1,(i-1))];
end
PP_shift

But its showing me error. Can anyone please help me? Thanks.

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Knut 2011-12-14

I dont have Matlab here, but could you not do something like:

[rows cols]= size(PP);

PP_shift = zeros(size(PP));

for i = 1:rows

PP_shift(i,1:end-i+1) = PP(i,i:end);

end

PP_shift

BTW, there is probably some neat 2d/1d way of indexing PP that will give you the desired result with faster execution. Perhaps ind2sub().

Ayesa 2011-12-14

Thank you so much. It works perfectly. Again many thanks.

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Answer 1

Walter Roberson 2011-12-14

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If "i" is the number of rows, then why are you using "i" to index the second dimension of B, which is the position for the column number?

Hint: B does not need to be a full-sized array; it could be a vector.

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Ayesa 2011-12-14

I use "i" to index the second dimension of B because for each row of PP_shift, there should be (i-1) column zeros of B inserted. So when rows = 1, B is a empty matrix, rows = 2, B is a 1 by 1 matrix ,when rows = 3, B is a 1 by 2 matrix and so on.

Otherwise how can i do that?

Walter Roberson 2011-12-14

Suppose PP is (say) 5 by 3. Then you would create B as 5 by 3 because you create it the same size as PP. Then when you got to row 5 (i=5) you would be trying to index B at (1,i-1) would would be (1,4) which would not exist because B would only be 3 wide.

hint: in your PP_shift line, replace B by the word zeros

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