## Check if subsequent elements in a 3D matrix are the same value

### mashtine (view profile)

on 28 Aug 2015
Latest activity Commented on by mashtine

on 28 Aug 2015

### Stephen Cobeldick (view profile)

Hello,
I have a 121x97x51135 matrix of 1's and 0's which flag certain periods of time. I would like to produce new variables that count the number of times a 1 is followed by another in the subsequent element, and another if two ones follow in the subsequent 2 elements.
For instance:
if
A(:,:,1) = ones(3,3,1);
A(:,:,2) = ones(3,3,1);
A(:,:,3) = zeros(3,3,1);
A(:,:,4) = ones(3,3,1);
The output variable for the first test (1 followed by 1) would be a 3x3 matrix with a count of the number of times that occurred in A and thus result = ones(3,3).
Any suggestions how to do this with out a loop? Not sure if a function out there works for this. I tried diff() but diff won't work in terms of adding up these times. For instance:
diff(A,[],3)
ans(:,:,1) =
0 0 0
0 0 0
0 0 0
ans(:,:,2) =
-1 -1 -1
-1 -1 -1
-1 -1 -1
ans(:,:,3) =
1 1 1
1 1 1
1 1 1

#### 1 Comment

Stephen Cobeldick

### Stephen Cobeldick (view profile)

on 28 Aug 2015
Giving us the input values A to try is great, but we also need to know what your desired output data is. Showing us the not working diff is not so useful, because although it is in your mind it does not tell us what the correct answer should be. Please give exact example output values!
How many "tests" do you need to run?

### Stephen Cobeldick (view profile)

on 28 Aug 2015
Edited by Stephen Cobeldick

### Stephen Cobeldick (view profile)

on 28 Aug 2015

>> Y2 = X(:,:,1:end-1) & X(:,:,2:end);
>> Y3 = X(:,:,1:end-2) & X(:,:,2:end-1) & X(:,:,3:end);
>> sum(Y2,3) % two consecutive "ones"
ans =
1 1 1
1 1 1
1 1 1
>> sum(Y3,3) % three consecutive "ones"
ans =
0 0 0
0 0 0
0 0 0

mashtine

### mashtine (view profile)

on 28 Aug 2015
Thanks for this Stephen! I would have never have thought this was possible nor know how to find this. Very good use of logical indexing!