How to fastly calculate this matrix operation

A matrix μ is a dimensional matrix, p is a dimensional vector, q is a dimensional matrix, c is a constant.

How to fastly calculate the following three dimensional () matrix X, where the element in X is

[Rik]

Let's start at the beginning: how would you write the slow code that does this? You can even start with random data:

K=10;
J=12;
I=14;
c=rand;
mu = rand(K,J);
p = rand(K,1);
q = rand(K,I);

[John D'Errico]

Unless I==J, then your expression does not even make mathematical sense. That is, you tell us that q is KxI, but then you index it as q(r,j), where r varies from 1 to K. Even then, you seem to be confusing I and J.

Next, the numerator has the shape of a vector, possibly of length either J or I, this is not clear. But the denominator is a 3 dimensional thing.

I would STRONGLY suggest that you need to use different variables than I and J here, as I think even you have become confused in what you wrote yourself.

Honestly, I'm not even sure how I would write that were I to use loops.

[Torsten]

@John D'Errico

That is, you tell us that q is KxI, but then you index it as q(r,j), where r varies from 1 to K.

It's indexed q(r,i).

Next, the numerator has the shape of a vector, possibly of length either J or I, this is not clear. But the denominator is a 3 dimensional thing.

I think numerator and denominator are just scalars.

[John D'Errico]

Sorry. Bad eyesight on my part. it is q(r,i). I thought that was a j. Again, the problem with using I and J. is they are easily confused.

As far as the denominator being a scalar, you can also view it as a 3-dimensional array, of the same shape as X. And I think, as you (@Torsten) knows, that is how you would perform the computation.

Answer 1

[John D'Errico]

First, use better variables. I is a terrible name to use , especially capital I, since it is so easily confused with the number 1, and even a lower case L (l). Depending on the font, they can be indistinguishable.

So I will assume that U is a matrix of size Kx1xM. The 1 there is important. And if you have created U to be of size KxN, then you need to reshape it so it is the size I show.

Assume p is a column vector, of length K, so in MATLAB, implicitly of size Kx1x1.

Assume q is an array, of size KxN. In MATLAB, such a 2-d array is also implicitly of size KxNx1.

I'll pick some random arrays.

K = 3;
N = 4;
M = 5; % all arbitrary, just to make an example.

u = rand(K,1,M);
p = rand(K,1,1);
q = rand(K,N,1);

c = randn(); % I'll pick some value for c too

First, how would you compute the denominator?

den = u.^2.*p.*q;

That was trivially done, since MATLAB is smart enough to expand the singleton dimensions into 3-dimensional arrays.

How about the numerator?

num = c + sum(u.^2.*p.*(1-q),1); % the sum will be performed over the first dimension.

Now I would note that num is an array, of size 1xNxM.

X = num./den

X = 
X(:,:,1) =

   1.0e+03 *

    2.5730    0.3042    0.2057    0.3599
    0.0013    0.0115    0.0181    0.0079
    0.0002    0.0010    0.0015    0.0046


X(:,:,2) =

  192.3316   24.7216   16.0489   25.4394
    0.6655    6.5974    9.9161    3.9023
    0.2438    1.1346    1.6771    4.6834


X(:,:,3) =

   96.8484   14.0472    9.1383   15.2627
    0.5831    6.5235    9.8254    4.0742
    0.1876    0.9850    1.4589    4.2928


X(:,:,4) =

 -781.9146  -26.1164  -13.5545  -15.1076
  -14.9784  -38.5860  -46.3653  -12.8301
   -2.0081   -2.4281   -2.8693   -5.6342


X(:,:,5) =

   49.6421    1.8893    0.7914    2.0345
   42.2240  123.9404  120.2054   76.7182
    0.7737    1.0659    1.0167    4.6044

The only thing I would question is if you intended the sum to be from 1 to K or from 1 to k. And that would be a difference. We would use cumsum in the latter case.

[Hancheng Zhu]

great, bro! You have solved this problem accurately. The only question is that , for example, u is provided as a two-dimensional matrix (). How to fastly scaled it into a three-dimensional matrix ()? I think a possible way is just using this code , but i don't know if there is other way that much faster than this one

[Torsten]

Note that the size of X is (K x N x M), not (K x M x N). Thus again some transposing is required.

[John D'Errico]

NO. Don't copy the columns over one at a tie. USE RESHAPE.

help reshape

 RESHAPE Reshape array by rearranging existing elements
    B = RESHAPE(A,M,N) or RESHAPE(A,[M,N]) returns the M-by-N matrix whose
    elements are taken columnwise from A. A must have M*N elements.
 
    B = RESHAPE(A,M,N,P,...) or RESHAPE(A,[M,N,P,...]) returns a
    multidimensional array with the same elements as A but reshaped to have
    the size M-by-N-by-P-by-.... The product of the specified dimensions,
    M*N*P*..., must be the same as NUMEL(A).
 
    B = RESHAPE(A,...,[],...) reshapes A according to the specified
    dimension sizes. You can specify a single dimension size of [] to have
    the dimension size automatically calculated, such that the product of
    the dimensions equals NUMEL(A). The value of NUMEL(A) must be evenly
    divisible by the product of the specified dimensions.
 
    See also SQUEEZE, SHIFTDIM, COLON, RESIZE.

    Documentation for reshape
       doc reshape

    Other uses of reshape

       categorical/reshape
       codistributed/reshape
       datetime/reshape
       digraph/reshape
       dlarray/reshape
       duration/reshape
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       matlab.mixin.indexing.RedefinesParen/reshape
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       se3/reshape
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       so3/reshape
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       tokenizedDocument/reshape

u = rand(5,3)

u = 5x3

0.2219 0.7364 0.4874 0.3214 0.6192 0.0274 0.4354 0.0881 0.8771 0.0893 0.0874 0.4635 0.3561 0.9840 0.7098

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% convert to a 5x1x3
uhat = reshape(u,size(u,1),1,size(u,2))

uhat = 
uhat(:,:,1) =

    0.2219
    0.3214
    0.4354
    0.0893
    0.3561


uhat(:,:,2) =

    0.7364
    0.6192
    0.0881
    0.0874
    0.9840


uhat(:,:,3) =

    0.4874
    0.0274
    0.8771
    0.4635
    0.7098

size(uhat)

ans = 1x3

5 1 3

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[Hancheng Zhu]

Thanks bro!