Sum if conditions are satisfied across arrays

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Abhishek Varghese 2018-2-14

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https://ww2.mathworks.cn/matlabcentral/answers/382597-sum-if-conditions-are-satisfied-across-arrays

编辑： Abhishek Varghese 2018-2-14

采纳的回答： Matt J

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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.

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Matt J 2018-2-14

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 2018-2-14

That's right. Edited it now. Thanks

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Answer 1

Matt J 2018-2-14

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/382597-sum-if-conditions-are-satisfied-across-arrays#answer_305080

编辑：Matt J 2018-2-14

在 MATLAB Online 中打开

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

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Matt J 2018-2-14

在 MATLAB Online 中打开

In that case,

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

Abhishek Varghese 2018-2-14

编辑：Abhishek Varghese 2018-2-14

在 MATLAB Online 中打开

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.

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