Once again, this is a situation that would be trivial if you had the Image Processing Toolbox because in one line of code you can find the lengths of all non-zero stretches of rainfall for a column. One more line of code would tell you the longest of those stretches. Do you have that toolbox? Type ver to find out. If you do, try this code:
a=[99 99 0.100 0.120 0 0.500 0;
0 0 0 0 0.150 0.120 0;
0.110 0.010 0.010 0 0 0.300 0.100;
0 0.250 0 0.050 0.060 0.100 0.110;
0 0.120 0.040 0 0.500 0.750 99]
[rows columns] = size(a);
for row = 1 : rows
measurements = regionprops(a(row, :)< 99 & a(row, :) > 0, 'Area');
% Get all the stretches of rain
allRains = [measurements.Area]
% Get the longest rain.
longestStretch = max(allRains);
fprintf('For row %d, the longest stretch with rain = %d.\n\n', ...
row, longestStretch);
end
In the command window:
allRains =
2 1
For row 1, the longest stretch with rain = 2.
allRains =
2
For row 2, the longest stretch with rain = 2.
allRains =
3 2
For row 3, the longest stretch with rain = 3.
allRains =
1 4
For row 4, the longest stretch with rain = 4.
allRains =
2 2
For row 5, the longest stretch with rain = 2.
