element wise multiplication and sum

hi,

i have a matirx a = [1,2,3;4,5,6]; and another matrix b=[2,2,2]

i want to multiply a[i,:].*b where i=1,2 (i.e. the rows of matrix a).

this will result in a matrix y with two rows (with 15 in row1 and 30 in row2)

how do i achieve this efficently without writing a "for" loop.

thanks in advance

[Azzi Abdelmalek]

What is the expected result?

Answer 1

[Matt Fig]

I don't know where the 15 came from. I assume you mean 12.

>> a = [1,2,3;4,5,6];
>> b = [2,2,2];
>> a*b.'   % same results as:  sum(a.*[b;b],2)

ans =

      12
      30

[Evan]

Oh wow, I can't believe I missed that. Yes, this is a much better answer. :P

[Yency Perez]

Hi, i'm wondering how to do it using a "for" loop!

Answer 2

[Akbar Khan]

As per my understanding of internal implementation of matlab. Matrix multiplication and matrix addition is an O(n^3) and O(n^2) time complexity algorithm. However I am not sure whether Strassen's algorithm is implemented internally.

Answer 3

[Evan]

NOTE: Ignore my answer. Matt's is much better. :P

>> a = [1,2,3;4,5,6]; >> b = [2,2,2]; >> tic >> c = bsxfun(@times,a,b) >> toc

   c =  
        2     4     6
        8    10    12
   Elapsed time is 0.000333 seconds.

For more info:

help bsxfun

Another (faster) way would be to resize b to be the same size as a and then perform element-wise multiplication. So something like this:

>> a = [1,2,3;4,5,6];
>> b = [2,2,2];
>> n = 2; %OR n = size(a,1);
>> tic
>> c = a .* b(ones(n,1),:) %second term is same as (ones(n,1) * b)
>> toc

    c =  
         2     4     6
         8    10    12
    Elapsed time is 0.000040 seconds.

You would just have to find the number of rows in a and use that as n for a general case.

Then, of course, you would just use the "sum" function to get your answer. I'm guessing you meant you wanted your answer to be [12; 30]? If so, you would just sum along columns:

>> y = sum(c,2)

    y =
        12
        30