Find amount of nonzero elements in matrix

Hello all,

I have a matrix that contains two column. Column A is the time and column B is the value belonging to that time. Now, column B contains zero and nonzero elements. What I need to do is count all the zeros and nonzeros. So I choose for a logical that returns true if it is nonzero and false if it is zero. I need to count the amount of consecutive nonzero elements because this corresponds to the duration. I used the following code to do this:

CntsCheckDataD1 = CheckDataD1(:,2);
        CheckDataD1log = CntsCheckDataD1 ~= 0;
        CheckDataD1log = CheckDataD1log';
        if (CheckDataD1log(1,1) == 0) == 1
            CheckDataD1log = diff([0 find(diff(CheckDataD1log)) numel(CheckDataD1log)]);
            CheckDataD1log = CheckDataD1log';
            onesCheckDataD1log = CheckDataD1log(2:2:end, :);
        else
            CheckDataD1log = diff([0 find(diff(CheckDataD1log)) numel(CheckDataD1log)]);
            CheckDataD1log = CheckDataD1log';
            onesCheckDataD1log = CheckDataD1log(1:2:end, :)

end

This returns all amount of ones in the matrix. I need to check where the first amount of consecutive ones is more than (for example) 15 and return the corresponding time from the CheckDataD1 matrix. Finding this value in the onesCheckDataD1log is easy but I do not know how I can find the corresponding time belonging to that value in the CheckDataD1 matrix..

Please help!

Answer 1

[Ameer Hamza]

Try regionprops(). It will tell you exactly how many times there are consecutive zeros groups, along with length and location of each group.

x = [ 1 0 0 0 1 2 0 0 0 1  1 0 0];
groups = regionprops(x==0)
groups = 
3×1 struct array with fields:  % <--- Indicating there are 3 groups of consecutive zeros
  Area
  Centroid
  BoundingBox

To get the length of each group

[groups.Area]
ans =
   3     3     2   % <---- Length of each group of zero, first group have 3 zeros, second has 3 and last have 2, same as in vector x

To get the location of each group

ceil(vertcat(groups.BoundingBox))
ans =
   2     1     3     1
   7     1     3     1
  12     1     2     1

ignore the 2nd and 4 column. The first column indicates the starting position and the third column indicate the length. E.g. first row [2 1 3 1] indicate that first group start at location 2 and have three zeros.

[Guillermo]

It works perfect, Thi is great!!! Thanks.

Answer 2

[KSSV]

Read about nnz. This gives you number of non-zeros in the data.

https://in.mathworks.com/help/matlab/ref/nnz.html

[Debbie Oomen]

Yes, I have. However this is not what I am looking for. I need to know the exact time when there is a specific amount of consecutive non-zero elements.

[KSSV]

You may follow like this:

A = [0,0,0,0,3,2,5,2,4,1,5,0,1,2,5,0,0,0,0,0,0,0,0,0,1,1,2,6,3,1];
t = 1:length(A) ;
ii = zeros(size(A));
jj = A > 0;
ii(strfind([0,jj(:)'],[0 1])) = 1;
idx = cumsum(ii).*jj;
x = accumarray( idx(jj)',t(jj)',[],@(x){x'});
y = accumarray( idx(jj)',A(jj)',[],@(x){x'});

for i = 1:length(x)
    [x{i}' y{i}']
end

Code taken from: https://in.mathworks.com/matlabcentral/answers/104614-grouping-continuous-nonzero-in-a-row-vector

[Debbie Oomen]

This gives me the following error:

Second input VAL must be a vector with one element for each row in SUBS, or a scalar.

[KSSV]

How did you try the code? I should see the code, you tried..along with data.

Answer 3

[Jan]

It sounds like FEX: RunLength (link) would help:

[Value, Repetitions, Index] = RunLength(CheckDataD1log)

[Debbie Oomen]

It says that it is an undefined function? And it does not necessarily have to be the largest amount of nonzero elements. So will this work then?

[Stephen23]

@Debbie Oomen: you need to download it from the link that Jan Simon gave you.