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Octave: How to sum(A .* B, 3) without expanding A .* B?

How do I create a simple Octave distributable without installing OctaveSumming diagonal elements of matrix in Octaveexpand, factor and simplify on OctaveGradient Descent implementation in octaveDifferences between Octave and MATLAB?Summing a single row vector by column in OctaveBasic for loop in Octave expandedSum of anonymous functions in OctaveHow to prevent Octave from summing row and column vectors?sum elements of vector excluding the first element in Octave

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2

Consider the following scenario, for A with size [k, 1, m] and B with size [1, n, m], how can one get the same result as:

C = sum(A .* B, 3);

without expanding

A .* B

Because that takes way too much memory.
Something like the following loop but natively:

C = zeros(k,n);
for idx = 1:m
 C += A(:,1,idx) * B(1,:,idx);
end

I guess I could also ask if there's a function like bsxfun with a "reduce"-like behavior?
Something like:

C = bsxfun_accumulate(@(a, b) a * b, A, B);

Note: by native I mean cs/cuda code-paths, or opencl code-path, or x86-sse, or plain x86 instructions. Whatever is available.

octave

share|improve this question

edited Mar 27 at 17:02

hackdev

asked Mar 27 at 16:43

hackdevhackdev

30822 silver badges77 bronze badges

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  What do you mean by "but natively"? Loops are native to MATLAB, and not necessarily slow. You just need to use C = C + ... rather than += which isn't valid MATLAB syntax.
  
  – Wolfie
  Mar 27 at 16:46
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  In addition to what Wolfie said, you also want to use * (matrix multiplication) instead of .* (element-wise multiplication) in your for loop.
  
  – gnovice
  Mar 27 at 16:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  By native I mean machine code. x86 or gpu code, definitely not interpreted. I changed the title to Octave only as my impression is that Matlab was much slower than octave. Is Matlab running on jvm or machine code?
  
  – hackdev
  Mar 27 at 16:55
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @hackdev I doubt MATLAB is slower than octave. Recent JIT compilers of MATLAB make it run as fast as native code. Also most of the operations (such as .*) are not run in MATLAB code, but in the underlying Fotran accelerated multi-threaded inbuilt function. MATLAB is fast for matrix operations, almost unbeatable fast. It also has the option of running most of the basic operations on the GPU.
  
  – Ander Biguri
  Mar 27 at 17:02
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  If Octave was faster than MATLAB, The MathWorks would go out of business... I don't understand why this is your assumption?
  
  – Wolfie
  Mar 27 at 17:06
  
  

  
  
  
  

 | 
show 1 more comment

2

Consider the following scenario, for A with size [k, 1, m] and B with size [1, n, m], how can one get the same result as:

C = sum(A .* B, 3);

without expanding

A .* B

Because that takes way too much memory.
Something like the following loop but natively:

C = zeros(k,n);
for idx = 1:m
 C += A(:,1,idx) * B(1,:,idx);
end

I guess I could also ask if there's a function like bsxfun with a "reduce"-like behavior?
Something like:

C = bsxfun_accumulate(@(a, b) a * b, A, B);

Note: by native I mean cs/cuda code-paths, or opencl code-path, or x86-sse, or plain x86 instructions. Whatever is available.

octave

share|improve this question

edited Mar 27 at 17:02

hackdev

asked Mar 27 at 16:43

hackdevhackdev

30822 silver badges77 bronze badges

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  What do you mean by "but natively"? Loops are native to MATLAB, and not necessarily slow. You just need to use C = C + ... rather than += which isn't valid MATLAB syntax.
  
  – Wolfie
  Mar 27 at 16:46
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  In addition to what Wolfie said, you also want to use * (matrix multiplication) instead of .* (element-wise multiplication) in your for loop.
  
  – gnovice
  Mar 27 at 16:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  By native I mean machine code. x86 or gpu code, definitely not interpreted. I changed the title to Octave only as my impression is that Matlab was much slower than octave. Is Matlab running on jvm or machine code?
  
  – hackdev
  Mar 27 at 16:55
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  @hackdev I doubt MATLAB is slower than octave. Recent JIT compilers of MATLAB make it run as fast as native code. Also most of the operations (such as .*) are not run in MATLAB code, but in the underlying Fotran accelerated multi-threaded inbuilt function. MATLAB is fast for matrix operations, almost unbeatable fast. It also has the option of running most of the basic operations on the GPU.
  
  – Ander Biguri
  Mar 27 at 17:02
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  If Octave was faster than MATLAB, The MathWorks would go out of business... I don't understand why this is your assumption?
  
  – Wolfie
  Mar 27 at 17:06
  
  

  
  
  
  

 | 
show 1 more comment

Consider the following scenario, for A with size [k, 1, m] and B with size [1, n, m], how can one get the same result as:

C = sum(A .* B, 3);

without expanding

A .* B

Because that takes way too much memory.
Something like the following loop but natively:

C = zeros(k,n);
for idx = 1:m
 C += A(:,1,idx) * B(1,:,idx);
end

I guess I could also ask if there's a function like bsxfun with a "reduce"-like behavior?
Something like:

C = bsxfun_accumulate(@(a, b) a * b, A, B);

Note: by native I mean cs/cuda code-paths, or opencl code-path, or x86-sse, or plain x86 instructions. Whatever is available.

octave

share|improve this question

edited Mar 27 at 17:02

hackdev

asked Mar 27 at 16:43

hackdevhackdev

30822 silver badges77 bronze badges

C = sum(A .* B, 3);

A .* B

C = zeros(k,n);
for idx = 1:m
 C += A(:,1,idx) * B(1,:,idx);
end

C = bsxfun_accumulate(@(a, b) a * b, A, B);

share|improve this question

edited Mar 27 at 17:02

hackdev

asked Mar 27 at 16:43

hackdevhackdev

30822 silver badges77 bronze badges

share|improve this question

edited Mar 27 at 17:02

hackdev

asked Mar 27 at 16:43

hackdevhackdev

30822 silver badges77 bronze badges

asked Mar 27 at 16:43

hackdevhackdev

30822 silver badges77 bronze badges

asked Mar 27 at 16:43

hackdevhackdev

30822 silver badges77 bronze badges

1 Answer
1

active

oldest

votes

6

You can actually solve your problem by simply reshaping the variables A and B and applying a matrix multiply:

C = reshape(A, [], m)*(reshape(B, [], m).');

Basically, summing the results of m sets of multiplications involving k-by-1 column vectors and 1-by-n row vectors is the equivalent of multiplying a k-by-m matrix of your columns and an m-by-n matrix of your rows.

share|improve this answer

edited Mar 27 at 17:23

answered Mar 27 at 17:02

gnovicegnovice

120k1313 gold badges237237 silver badges341341 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Gee, that not only seems to work, it's actually ridiculously fast compared to the loop. I would give you two votes, but I can only vote once. XD
  
  – hackdev
  Mar 27 at 17:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @hackdev as asker of the question, you can mark the answer as "accepted", thereby closing the question and awarding 1.5x the reputation points of an upvote.
  
  – Wolfie
  Mar 27 at 17:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I was having issues with this question because I have a corner case where A is [k, 1, 1], where matrix multiplication doesn't work. So I realized that .* always works. But in practice it explodes for large matrices. Basically adding an if size(A,3)==1 solves the whole issue. :)
  
  – hackdev
  Mar 27 at 17:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @hackdev: If you are faced with a case where A is size [k 1 1] but B is still size [1 n m], here's one way to solve it like above: C = repmat(A, 1, m)*(reshape(B, [], m).');
  
  – gnovice
  Mar 27 at 17:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That last corner case works well by using A * sum(B'), but I'll measure to see which is faster.
  
  – hackdev
  Mar 27 at 17:31
  
  

  
  
  
  

 | 
show 2 more comments

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6

You can actually solve your problem by simply reshaping the variables A and B and applying a matrix multiply:

C = reshape(A, [], m)*(reshape(B, [], m).');

Basically, summing the results of m sets of multiplications involving k-by-1 column vectors and 1-by-n row vectors is the equivalent of multiplying a k-by-m matrix of your columns and an m-by-n matrix of your rows.

share|improve this answer

edited Mar 27 at 17:23

answered Mar 27 at 17:02

gnovicegnovice

120k1313 gold badges237237 silver badges341341 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Gee, that not only seems to work, it's actually ridiculously fast compared to the loop. I would give you two votes, but I can only vote once. XD
  
  – hackdev
  Mar 27 at 17:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @hackdev as asker of the question, you can mark the answer as "accepted", thereby closing the question and awarding 1.5x the reputation points of an upvote.
  
  – Wolfie
  Mar 27 at 17:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I was having issues with this question because I have a corner case where A is [k, 1, 1], where matrix multiplication doesn't work. So I realized that .* always works. But in practice it explodes for large matrices. Basically adding an if size(A,3)==1 solves the whole issue. :)
  
  – hackdev
  Mar 27 at 17:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @hackdev: If you are faced with a case where A is size [k 1 1] but B is still size [1 n m], here's one way to solve it like above: C = repmat(A, 1, m)*(reshape(B, [], m).');
  
  – gnovice
  Mar 27 at 17:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That last corner case works well by using A * sum(B'), but I'll measure to see which is faster.
  
  – hackdev
  Mar 27 at 17:31
  
  

  
  
  
  

 | 
show 2 more comments

6

You can actually solve your problem by simply reshaping the variables A and B and applying a matrix multiply:

C = reshape(A, [], m)*(reshape(B, [], m).');

Basically, summing the results of m sets of multiplications involving k-by-1 column vectors and 1-by-n row vectors is the equivalent of multiplying a k-by-m matrix of your columns and an m-by-n matrix of your rows.

share|improve this answer

edited Mar 27 at 17:23

answered Mar 27 at 17:02

gnovicegnovice

120k1313 gold badges237237 silver badges341341 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Gee, that not only seems to work, it's actually ridiculously fast compared to the loop. I would give you two votes, but I can only vote once. XD
  
  – hackdev
  Mar 27 at 17:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @hackdev as asker of the question, you can mark the answer as "accepted", thereby closing the question and awarding 1.5x the reputation points of an upvote.
  
  – Wolfie
  Mar 27 at 17:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I was having issues with this question because I have a corner case where A is [k, 1, 1], where matrix multiplication doesn't work. So I realized that .* always works. But in practice it explodes for large matrices. Basically adding an if size(A,3)==1 solves the whole issue. :)
  
  – hackdev
  Mar 27 at 17:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @hackdev: If you are faced with a case where A is size [k 1 1] but B is still size [1 n m], here's one way to solve it like above: C = repmat(A, 1, m)*(reshape(B, [], m).');
  
  – gnovice
  Mar 27 at 17:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That last corner case works well by using A * sum(B'), but I'll measure to see which is faster.
  
  – hackdev
  Mar 27 at 17:31
  
  

  
  
  
  

 | 
show 2 more comments

You can actually solve your problem by simply reshaping the variables A and B and applying a matrix multiply:

C = reshape(A, [], m)*(reshape(B, [], m).');

Basically, summing the results of m sets of multiplications involving k-by-1 column vectors and 1-by-n row vectors is the equivalent of multiplying a k-by-m matrix of your columns and an m-by-n matrix of your rows.

share|improve this answer

edited Mar 27 at 17:23

answered Mar 27 at 17:02

gnovicegnovice

120k1313 gold badges237237 silver badges341341 bronze badges

C = reshape(A, [], m)*(reshape(B, [], m).');

share|improve this answer

edited Mar 27 at 17:23

answered Mar 27 at 17:02

gnovicegnovice

120k1313 gold badges237237 silver badges341341 bronze badges

answered Mar 27 at 17:02

gnovicegnovice

120k1313 gold badges237237 silver badges341341 bronze badges

answered Mar 27 at 17:02

gnovicegnovice

120k1313 gold badges237237 silver badges341341 bronze badges