Subtract matrices of different sizes











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I have $2$ matrices of different sizes: $8 times 3$ matrix $A$ and $2 times 3$ matrix $B$. I need to subtract from each row of $B$, a particular row of $A$ and sum these values. This I need to do for all the rows for $A$ and store the result in another matrix $C$. Is there a way to do this avoiding both for loops?



I can avoid one using bsxfun but seems like I need to use one for loop for iterating over the elements of $A$.










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    up vote
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    I have $2$ matrices of different sizes: $8 times 3$ matrix $A$ and $2 times 3$ matrix $B$. I need to subtract from each row of $B$, a particular row of $A$ and sum these values. This I need to do for all the rows for $A$ and store the result in another matrix $C$. Is there a way to do this avoiding both for loops?



    I can avoid one using bsxfun but seems like I need to use one for loop for iterating over the elements of $A$.










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I have $2$ matrices of different sizes: $8 times 3$ matrix $A$ and $2 times 3$ matrix $B$. I need to subtract from each row of $B$, a particular row of $A$ and sum these values. This I need to do for all the rows for $A$ and store the result in another matrix $C$. Is there a way to do this avoiding both for loops?



      I can avoid one using bsxfun but seems like I need to use one for loop for iterating over the elements of $A$.










      share|cite|improve this question















      I have $2$ matrices of different sizes: $8 times 3$ matrix $A$ and $2 times 3$ matrix $B$. I need to subtract from each row of $B$, a particular row of $A$ and sum these values. This I need to do for all the rows for $A$ and store the result in another matrix $C$. Is there a way to do this avoiding both for loops?



      I can avoid one using bsxfun but seems like I need to use one for loop for iterating over the elements of $A$.







      matrices matlab






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      edited Apr 1 at 10:05









      Rodrigo de Azevedo

      12.8k41853




      12.8k41853










      asked Mar 11 '14 at 19:50









      Shrinu Kushagra

      61




      61






















          2 Answers
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          If I understand correctly, you want to (1) subtract every row of A from every row of B, and then (2) sum all values within each resulting row.





          1. For step (1) you can use a single bsxfun, but you need to permute dimensions:



            subtractedRows = bsxfun(@minus, B, permute(A,[3 2 1]));


            For any m, n, subtractedRows(m,:,n) gives B(m,:)-A(n,:).




          2. For step (2) you only need to sum along second dimension:



            result = squeeze(sum(subtractedRows,2));







          share|cite|improve this answer




























            up vote
            0
            down vote













            It sounds like you can sum across the columns of A and B before subtracting and simplify the computation.



            If broadcasting is available (newest versions of Matlab and Octave):



            C = sum(B, 2) - sum(A, 2)'


            If not:



            C = repmat(sum(B, 2), [1 rows(A)]) - repmat(sum(A, 2)', [rows(B) 1])





            share|cite|improve this answer





















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              2 Answers
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              active

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              2 Answers
              2






              active

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              active

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              up vote
              0
              down vote













              If I understand correctly, you want to (1) subtract every row of A from every row of B, and then (2) sum all values within each resulting row.





              1. For step (1) you can use a single bsxfun, but you need to permute dimensions:



                subtractedRows = bsxfun(@minus, B, permute(A,[3 2 1]));


                For any m, n, subtractedRows(m,:,n) gives B(m,:)-A(n,:).




              2. For step (2) you only need to sum along second dimension:



                result = squeeze(sum(subtractedRows,2));







              share|cite|improve this answer

























                up vote
                0
                down vote













                If I understand correctly, you want to (1) subtract every row of A from every row of B, and then (2) sum all values within each resulting row.





                1. For step (1) you can use a single bsxfun, but you need to permute dimensions:



                  subtractedRows = bsxfun(@minus, B, permute(A,[3 2 1]));


                  For any m, n, subtractedRows(m,:,n) gives B(m,:)-A(n,:).




                2. For step (2) you only need to sum along second dimension:



                  result = squeeze(sum(subtractedRows,2));







                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  If I understand correctly, you want to (1) subtract every row of A from every row of B, and then (2) sum all values within each resulting row.





                  1. For step (1) you can use a single bsxfun, but you need to permute dimensions:



                    subtractedRows = bsxfun(@minus, B, permute(A,[3 2 1]));


                    For any m, n, subtractedRows(m,:,n) gives B(m,:)-A(n,:).




                  2. For step (2) you only need to sum along second dimension:



                    result = squeeze(sum(subtractedRows,2));







                  share|cite|improve this answer












                  If I understand correctly, you want to (1) subtract every row of A from every row of B, and then (2) sum all values within each resulting row.





                  1. For step (1) you can use a single bsxfun, but you need to permute dimensions:



                    subtractedRows = bsxfun(@minus, B, permute(A,[3 2 1]));


                    For any m, n, subtractedRows(m,:,n) gives B(m,:)-A(n,:).




                  2. For step (2) you only need to sum along second dimension:



                    result = squeeze(sum(subtractedRows,2));








                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 14 '14 at 11:13









                  Luis Mendo

                  1,3321821




                  1,3321821






















                      up vote
                      0
                      down vote













                      It sounds like you can sum across the columns of A and B before subtracting and simplify the computation.



                      If broadcasting is available (newest versions of Matlab and Octave):



                      C = sum(B, 2) - sum(A, 2)'


                      If not:



                      C = repmat(sum(B, 2), [1 rows(A)]) - repmat(sum(A, 2)', [rows(B) 1])





                      share|cite|improve this answer

























                        up vote
                        0
                        down vote













                        It sounds like you can sum across the columns of A and B before subtracting and simplify the computation.



                        If broadcasting is available (newest versions of Matlab and Octave):



                        C = sum(B, 2) - sum(A, 2)'


                        If not:



                        C = repmat(sum(B, 2), [1 rows(A)]) - repmat(sum(A, 2)', [rows(B) 1])





                        share|cite|improve this answer























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          It sounds like you can sum across the columns of A and B before subtracting and simplify the computation.



                          If broadcasting is available (newest versions of Matlab and Octave):



                          C = sum(B, 2) - sum(A, 2)'


                          If not:



                          C = repmat(sum(B, 2), [1 rows(A)]) - repmat(sum(A, 2)', [rows(B) 1])





                          share|cite|improve this answer












                          It sounds like you can sum across the columns of A and B before subtracting and simplify the computation.



                          If broadcasting is available (newest versions of Matlab and Octave):



                          C = sum(B, 2) - sum(A, 2)'


                          If not:



                          C = repmat(sum(B, 2), [1 rows(A)]) - repmat(sum(A, 2)', [rows(B) 1])






                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Sep 15 '16 at 14:44









                          Nir

                          58327




                          58327






























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