occupancy detection from a column of zeros and ones

Hello All,

I have some data which is basically some 0s and 1s like below in one column:

1
1
1
0
1
0
0
0
1
0
1
1
1
1
1
1
0
0
1
1
1
1
1
0
0
0
1
0
1
1
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
1
1
1
1
1
1
1
1
1
0
1
0
0
1
0
1
0
1
1
1
0
0
1
0
0
0
1
1
0
0
1
0
1
1
1
1
0
1
1
1
1
0
0
1
0
0
0
0

These show occupancy status of a location. When it is zero it means that the place is unoccupied, and when it is 1 it means that the place is occupied. The problem is that these data are not accurate. So the location is occupied if and only if there are at least 3 consecutive 1s happening, and it is unoccupied it there are at least 3 consecutive zeros happening. Can you please tell me how I can identify occupancy and non-occupancy data using this criteria? Thanks, Masih

Answer 1

[Image Analyst]

Use bwareaopen() or bwareafilt(). Both are in the Image Processing Toolbox.

occupancy = [...
  1
  1
  1
  0
  1
  0
  0
  0
  1
  0
  1
  1
  1
  1
  1
  1
  0
  0
  1
  1
  1
  1
  1
  0
  0
  0
  1
  0
  1
  1
  0
  1
  1
  1
  1
  1
  1
  1
  1
  1
  1
  1
  1
  1
  1
  1
  1
  1
  1
  0
  0
  1
  1
  1
  1
  1
  1
  1
  1
  1
  0
  1
  0
  0
  1
  0
  1
  0
  1
  1
  1
  0
  0
  1
  0
  0
  0
  1
  1
  0
  0
  1
  0
  1
  1
  1
  1
  0
  1
  1
  1
  1
  0
  0
  1
  0
  0
  0
  0]
% Get rid of regions 2 or less, leaving occupied elements only.
occupancy = bwareaopen(occupancy, 3)
% The inverse is the non occupied elements.
nonOccupied = ~occupancy

[masih]

Actually the answer I am trying to get from matlab should look like this: It should tell me that from the first row to the fifth row the place is occupied (ignoring the zero in the fourth row because it wasn't continuing for 3 consecutive times) for then from sixth row to the 9th row the place is unoccupied and so on...any help is appreciated.

[Image Analyst]

In a way, it does. That's what occupancy is. You can pass it into find if you want to convert to indexes

indexes = find(occupancy)

If you want it to say exactly "Region n goes from i1 to i2" then use this code:

% Get rid of regions 2 or less, leaving occupied elements only.
occupancy = bwareaopen(occupancy, 3)
% The inverse is the non occupied elements.
nonOccupied = ~occupancy
indexes = find(occupancy)
[bw, numRegions] = bwlabel(occupancy)
props = regionprops(bw, 'PixelList');
for p = 1 : numRegions
  thisRegion = props(p).PixelList(:, 2);
  fprintf('Region %d goes from element %d to element %d\n', ...
    p, thisRegion(1), thisRegion(end));
end

and you'll see it print out:

Region 1 goes from element 1 to element 3
Region 2 goes from element 11 to element 16
Region 3 goes from element 19 to element 23
Region 4 goes from element 32 to element 49
Region 5 goes from element 52 to element 60
Region 6 goes from element 69 to element 71
Region 7 goes from element 84 to element 87
Region 8 goes from element 89 to element 92

[masih]

Thanks a lot