Confusion regarding usage of "reshape"

Hi

During working on a project using Matlab R2018b, I came across a function, called "reshape". I try to understand how does it work and got stuck in 2 calculations where this dimesion matters.

Situation 1:

d=[1
0]
t=[val(:,:,1,1) = 0.0016
  val(:,:,2,1) =-4.7149e-04
  val(:,:,1,2) =-7.2994e-04
  val(:,:,2,2) =0.0027
]

I have tried to multiply d[1] with val(:,:,1,1) and val(:,:,2,1). And d[2] with val(:,:,1,2) and val(:,:,2,2)

I have written this code :

NewB = zeros(1,1,20,500);
NewB(i)=(d(i)*t(i));// i is for iteration through 4 th dimension of t

But I found that the my code multiplies only d[1] with val(:,:,1,1) and d[2] with val(:,:,1,2) only. I think , I need to alter the dimensions in calculation, right?

Situation 2:

A is of size [1 x 1 x 20 x 500] and B is of size [1 x 1 x 100 x 20]

I reshaped B following this:

B= reshape(B, 1,1,20,100);

Now I would like to multiply A[1 x 1 x 20 x1] with B[1x 1x 20 x1] and continue it for the remaining list.

Can you adivse me how to approach this problem?

thanks,

[Greg Heath]

Clarify your notation so that the interpretation of the variable x and the times symbol "x" will not get confused

Hope this helps

Greg

[Saugata Bose]

@Greg

thanks. i have edited the question as per ur advise. hope it will help u to help me.

Answer 1

[Walter Roberson]

See bsxfun(), and https://blogs.mathworks.com/loren/2016/10/24/matlab-arithmetic-expands-in-r2016b/ implicit expansion .

Or repmat() so the values are the same size.

[Saugata Bose]

@Walter Roberson

Thank you for sharing me the tutorial.

I think repmat() will help me solving situation2. But what would be the way to solve solution 1 where I would like to multiply one value from d with val(:,:,1,1) and with val(:,:,2,1)?