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 $A$.
matrices matlab
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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 $A$.
matrices matlab
add a comment |
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 $A$.
matrices matlab
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 at 10:05
Rodrigo de Azevedo
12.8k41853
12.8k41853
asked Mar 11 '14 at 19:50
Shrinu Kushagra
61
61
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2 Answers
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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.
For step (1) you can use a single
bsxfun
, but you need topermute
dimensions:
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));
add a comment |
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])
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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
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.
For step (1) you can use a single
bsxfun
, but you need topermute
dimensions:
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));
add a comment |
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.
For step (1) you can use a single
bsxfun
, but you need topermute
dimensions:
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));
add a comment |
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.
For step (1) you can use a single
bsxfun
, but you need topermute
dimensions:
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));
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 topermute
dimensions:
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 Mendo
1,3321821
1,3321821
add a comment |
add a comment |
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])
add a comment |
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])
add a comment |
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])
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
Nir
58327
58327
add a comment |
add a comment |
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