Need to find how many times in the matrix the Value 1 goes to Value 2

I'm looking for a way to find the number of times a specific transition occurs between two values in a matrix A. Matrix A is 9x9 and has values ranging from 0 to 8. I need to find the number of times a value, say 5, is immediately followed by another value, say 2.

%This an example of matrix A
A=[1 3 4 0;4 2 3 1;1 2 3 1]                                                         

For this matrix, I would need to find multiple combinations

1 to 3

4 to 0

3 to 1

etc.

I have no clue where to go with this. Thanks for any help.

Edit: Right now, i have a 8x30 excel spreadsheet with data that will be entered into a matrix. I then need to find different combinations of entries, where one value follows another value, ie 0 to 0, 0 to 1,etc.

this is just part of the data set, and the missing row will be filled in. So in the first row, 0 to 0 would result in an output of 1.

[Walter Roberson]

Does row 2 column 1 immediately follow row 1 column 9?

[Jesse Valentin]

If i understand the question correctly, no. each row is a different data set for how many people were inline at a coffee shop every two minutes.

[Walter Roberson]

Hint: you can use accumarray. Or just loop.

[Jesse Valentin]

Which example in the accumarray documentation could I follow?

[Image Analyst]

"each row is a different data set for how many people were inline at a coffee shop every two minutes." So then A cannot be a matrix - it must be a cell array unless you always have the same number of people in the shop, which is highly doubtful. Matrices must always have the same number of columns, but with a cell array (vector) you can have one cell for each row, and a variable length vector inside each row.

A{1} = [1 3 4 0]; % 4 elements (customers)
A{2} = [4 2 3 1 5 3 5 2 1 2]; % 10 elements (customers)
A{3} = [1 2 3 1 4 1 0 2]; % 8 elements (customers)

I think it would be very instructive for you to read the FAQ on cell arrays: Click here for the FAQ on cell arrays

[Walter Roberson]

p = randi([0 9],50,1);
accumarray([p(1:25)+1, p(26:50)+1], 1, [10 10])

[Jesse Valentin]

So for this example, would p be the input arrays that Image Analyst showed for each row of data? and how would the accumarray code change if the array has 30 entries? would it be

accumarrau([p(1:15)+1], p(16:30)+1],1, [10,10])

what are the 1 and tens at the end of this code?

Thanks for the help. Haven't been taught enough matlab from my professors.

[Walter Roberson]

In this example, p shows one row that is 50 long, and this example is for transitions between the first and 26th point, 2nd and 27'th point, 3rd and 28th point... same as if you had done

reshape(p, 25, 2)

and were doing the transitions for those (two element) rows. It is taking transitions

p(1), p(1+25)
p(2), p(2+25)
p(3), p(3+25)

which you could also write as

idx = 1:50-25;
[p(idx), p(idx+25)]

Now, if you wanted to measure transitions between elements that were (say) 10 apart rather than 25 apart, you could write

idx = 1:50-10;
[p(idx), p(idx+10)]

which you could also write as

[p(1:end-10), p((1:end-10)+10)]

Hint Hint. (You don't need to go one row at a time) Hint Hint.

[10 10] is because there are 10 different states each of which can transfer to each other. In your actual probably there are only 9 different states so you would use [9 9] for your real problem.

[Jesse Valentin]

So if i wanted to take transitions between 1 and 2, 2 and 3, 3 and 4 and so on...

A{1}=[1 2 3 4 5 6...]%30 total data entries
idk=1:30-1; %30 total columns
[A{1}(idx), A{1}(idx+1)]

Would the last line be put into the accumarray command like this?

accumarray([A{1}(idx), A{1}(idx+1)], 1, [30 30])

Edit: Also, how would I choose which transition I wanted to find?

[Walter Roberson]

The [10 10] is not because of 10 columns: it is because of 10 different states, since I used randi([0 9]). The [10 10] is choosing the output array size. Every state needs the possibility to transition to every other state so if NS is number of states then use [NS NS] .

You could also just leave out the argument and let accumarray size the array for you. But suppose that the only time state 8 is seen is at the beginning of rows, that nothing ever transitions from a lower state to state 8. Then [something, 8]+1 never occurs in the input, and accumarray would not have any reason to create a column 8+1 in the output array. accumarray only creates a row or column if it or higher number appears in the input.

Remember to do the +1 by the way,

out = accumarray([A{1}(idx), A{1}(idx+1)] + 1, 1, [NS NS])

this is needed because you have 0 in the input, and the values in the first position must all be positive integers; so the data for 0 is in row/column 1, the data for 1 is in row/column 2, and so on.

The transition count from state N to state M would be at out(N+1,M+1)

[Walter Roberson]

Hint Hint: if you think about the index numbers more then you will find that you do not need to work row by row. Not every possible idx, idx+1 can happen, but the ones that cannot happen are easy to define and eliminate.

[Jesse Valentin]

so i am at this right now..

A{1}=[2 0 0 1 2 3 3 4 1 2 1 2 1 2 5 2 1 1 0 0 2 0 0 1 1 0 1 2 0 2];
idx = 1:30-1;
out=accumarray([A{1}(idx), A{1}(idx+1)] + 1,1, [10,10]);

Thanks for all the help, but don't worry about it. i'll just do the process by hand

Answer 1

[Image Analyst]

Here's one way:

% Define data
A{1} = [1 3 4 0]; % 4 elements (customers)
A{2} = [4 2 3 1 5 3 5 2 1 2]; % 10 elements (customers)
A{3} = [1 2 3 1 4 1 0 2]; % 8 elements (customers)
% Instantiate output large enough to handle all expected numbers.
output = zeros(60*24, 60*24); % 60*24 is the number of minutes in a day.
% Compute counts of each pair.
for k = 1 : length(A)
	thisCell = A{k}
	for col = 1 : length(thisCell) - 1
		startValue = thisCell(col);
		endValue = thisCell(col + 1);
		% Increment value.  Add 1 to row and column because counts for row 0 must go into row 1.
		output(startValue + 1, endValue + 1) = output(startValue + 1, endValue + 1) + 1;
	end
end
[rows, columns] = find(output);
% Crop to largest times
output = output(1 : max(rows), 1 : max(columns))