Find Consecutive numbers and print value in next collumn

Hi,

I have a table of 4 collumns with over a million rows. In collumn 4 I have a collumn that iis based on logic (when the temperature in collumn 3 exceeds a given value the entry in collumn 4 will be one and when it does not exceed it will be 0).

In collumn 5 I would like to have a count of the number of consectutive values placed in the row of the final 1.

I have used the following to get a vector of containing the consecutive counts, but I am unsure how to get them to print in collumn five at the end of each cosecutive set

  f = find(diff([0,exceedlogic,0]==1));
   p = f(1:2:end-1);  % Start indices
   y = f(2:2:end)-p;  % Consecutive ones’ counts

[David Hill]

A simple example showing a few rows would be helpful

[Patrick Lonergan]

The data set is rather large but I have taken some screen shots of the table. The table is in fact 5 collumns. \

As you can see collumn 5 is filled with 0 and 1 (logic), in collumn six I would like the consecutive count. For example in row 19 collumn 6 the value 1 would be inserted, while in row 247 collumn 6 should be the value of 2.

I hope that clears things up.

Answer 1

[Image Analyst]

Try this:

fprintf('Beginning to run %s.m ...\n', mfilename);
% Create sample data because none was attached to the question.
data = rand(200, 6);
data(:, 5) = data(:, 4) > 0.3;
data(:, 6) = 0
% Now we have our data and can begin.

%==================================================================
% Measure lengths of each run of 1s.
props = regionprops(logical(data(:, 5)), 'Area', 'PixelList');
% Assign area to the very last row of 1s.
for k = 1 : length(props)
	lastRow = props(k).PixelList(end, 2)
	data(lastRow, 6) = props(k).Area;
end
%==================================================================

fprintf('Done running %s.m.\n', mfilename);

Main code is between the ============== of course.

[Patrick Lonergan]

Thanks worked great, cheers

Answer 2

[David Hill]

Simple loop works

c=0;
f(:,6)=0;
for k=1:size(f,1)
  if f(k,5)==1
    c=c+1;
    f(k,6)=c;
  else
    c=0;
  end
end

[Patrick Lonergan]

Thanks