Subtract matrices of different sizes
$begingroup$
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
edited Apr 1 '18 at 10:05
Rodrigo de Azevedo
13k41958
asked Mar 11 '14 at 19:50
Shrinu KushagraShrinu Kushagra
61
$endgroup$
add a comment |
$begingroup$
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
edited Apr 1 '18 at 10:05
Rodrigo de Azevedo
13k41958
asked Mar 11 '14 at 19:50
Shrinu KushagraShrinu Kushagra
61
$endgroup$
add a comment |
$begingroup$
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
edited Apr 1 '18 at 10:05
Rodrigo de Azevedo
13k41958
asked Mar 11 '14 at 19:50
Shrinu KushagraShrinu Kushagra
61
$endgroup$
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
matrices matlab
edited Apr 1 '18 at 10:05
Rodrigo de Azevedo
13k41958
asked Mar 11 '14 at 19:50
Shrinu KushagraShrinu Kushagra
61
edited Apr 1 '18 at 10:05
Rodrigo de Azevedo
13k41958
asked Mar 11 '14 at 19:50
Shrinu KushagraShrinu Kushagra
61
edited Apr 1 '18 at 10:05
Rodrigo de Azevedo
13k41958
edited Apr 1 '18 at 10:05
Rodrigo de Azevedo
13k41958
edited Apr 1 '18 at 10:05
Rodrigo de Azevedo
13k41958
13k41958
asked Mar 11 '14 at 19:50
Shrinu KushagraShrinu Kushagra
61
asked Mar 11 '14 at 19:50
Shrinu KushagraShrinu Kushagra
61
asked Mar 11 '14 at 19:50
Shrinu KushagraShrinu Kushagra
61
61
add a comment |
add a comment |
2 Answers
2
active
oldest
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$begingroup$
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.
For step (1) you can use a single
bsxfun, but you need topermutedimensions:
subtractedRows = bsxfun(@minus, B, permute(A,[3 2 1]));
For any
m,n,subtractedRows(m,:,n)givesB(m,:)-A(n,:).
For step (2) you only need to sum along second dimension:
result = squeeze(sum(subtractedRows,2));
answered Mar 14 '14 at 11:13
Luis MendoLuis Mendo
1,3321821
$endgroup$
add a comment |
$begingroup$
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])
answered Sep 15 '16 at 14:44
NirNir
65837
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
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.
For step (1) you can use a single
bsxfun, but you need topermutedimensions:
subtractedRows = bsxfun(@minus, B, permute(A,[3 2 1]));
For any
m,n,subtractedRows(m,:,n)givesB(m,:)-A(n,:).
For step (2) you only need to sum along second dimension:
result = squeeze(sum(subtractedRows,2));
answered Mar 14 '14 at 11:13
Luis MendoLuis Mendo
1,3321821
$endgroup$
add a comment |
$begingroup$
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.
For step (1) you can use a single
bsxfun, but you need topermutedimensions:
subtractedRows = bsxfun(@minus, B, permute(A,[3 2 1]));
For any
m,n,subtractedRows(m,:,n)givesB(m,:)-A(n,:).
For step (2) you only need to sum along second dimension:
result = squeeze(sum(subtractedRows,2));
answered Mar 14 '14 at 11:13
Luis MendoLuis Mendo
1,3321821
$endgroup$
add a comment |
$begingroup$
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.
For step (1) you can use a single
bsxfun, but you need topermutedimensions:
subtractedRows = bsxfun(@minus, B, permute(A,[3 2 1]));
For any
m,n,subtractedRows(m,:,n)givesB(m,:)-A(n,:).
For step (2) you only need to sum along second dimension:
result = squeeze(sum(subtractedRows,2));
answered Mar 14 '14 at 11:13
Luis MendoLuis Mendo
1,3321821
$endgroup$
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.
For step (1) you can use a single
bsxfun, but you need topermutedimensions:
subtractedRows = bsxfun(@minus, B, permute(A,[3 2 1]));
For any
m,n,subtractedRows(m,:,n)givesB(m,:)-A(n,:).
For step (2) you only need to sum along second dimension:
result = squeeze(sum(subtractedRows,2));
answered Mar 14 '14 at 11:13
Luis MendoLuis Mendo
1,3321821
answered Mar 14 '14 at 11:13
Luis MendoLuis Mendo
1,3321821
answered Mar 14 '14 at 11:13
Luis MendoLuis Mendo
1,3321821
answered Mar 14 '14 at 11:13
Luis MendoLuis Mendo
1,3321821
1,3321821
add a comment |
add a comment |
$begingroup$
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])
answered Sep 15 '16 at 14:44
NirNir
65837
$endgroup$
add a comment |
$begingroup$
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])
answered Sep 15 '16 at 14:44
NirNir
65837
$endgroup$
add a comment |
$begingroup$
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])
answered Sep 15 '16 at 14:44
NirNir
65837
$endgroup$
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])
answered Sep 15 '16 at 14:44
NirNir
65837
answered Sep 15 '16 at 14:44
NirNir
65837
answered Sep 15 '16 at 14:44
NirNir
65837
answered Sep 15 '16 at 14:44
NirNir
65837
65837
add a comment |
add a comment |
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