Sum if conditions are satisfied across arrays

Hello everyone,

Kinda stuck. Scenario is:

I have a binary 2D array that is n rows and m columns.

I want to count how many times an element went from being a '1' in one row, to a '0' in the next row, and store it as a vector.

I have updated an illustration to help.

Would really appreciate an efficient and simple solution, cheers.

[Matt J]

I have a binary 2D array that is m rows and n columns.

According to your diagram, n is the number of rows, not columns.

[Abhishek Varghese]

That's right. Edited it now. Thanks

Answer 1

[Matt J]

 result = sum(diff(array,1,1)==-1  ,1)

[Abhishek Varghese]

Hey Matt, sorry I may not have explained it correctly. The result vector should really be of length(array)-1 since it's comparing between the rows.

I've attached an image to the question to help visualise it better! Really appreciate it.

[Matt J]

In that case,

   result = sum( diff(array,1,1)==-1  ,2)

[Abhishek Varghese]

Related Question:

Hey Matt, still currently optimising my code. Would love further help.

Considering that I have the adjacency matrix:

      adj =     [0  1  1  1
                 1  0  0  1
                 1  0  0  1
                 1  1  1  0];

and a binary matrix:

array = randi([0 1],4,10)

where the columns are each node in the network, and the rows for the state of the network at each time step. To clarify, if node 1 is '1' at time step 2, it would be like

array(2,1) = 1

What would I do, to create a vector of values that satisfied the same condition as my initial question, and also satisfy the condition of not having any infected neighbours in the same time step??

EDIT: An infection is considered a 1, no infection a 0

EDIT2: I have decided that perhaps its best to create another question. Please find the question here:

To receive credit for the answer! Cheers -Abhishek.