PLEASE HELP - how to COUNT occurances

Hello,

I have an array (sortedresults) which has 3 columns (time, land, ilok) and each contains integer values. In column 2 (land) I need to COUNT the number of times that integer changes from 3 to anything else EXCEPT zero, AND verify the value of column 3 (ilok) is either the value of 2 or 3. When both conditions are met I need to add 1 to my COUNT.

Thanks in advance!

[Star Strider]

Please clarify: In column 2 (land) I need to COUNT the number of times that integer changes from 3 to anything else EXCEPT zero...

Does the direction of the change matter, i.e. does that only mean the change from row k to row k+1 or from row k-1 to row k, either, or both?

Gear-up landings can really take the fun out of your day! (I had to drop the gear by hand once. Fortunately it locked down.)

[steve]

CLARIFICATION - as you guessed this is a landing gear position signal and a ground interlock signal off an AC. I need to know when -

land(k) = 3 AND land(k+1) = 1,2, or 4...AND

when the above is true ilok(k) = 2 or 3

anytime both of those conditions are met I need to "count" that occurance.

Sorry, total newbie to MATLAB, last time I programmed was in F77, haha. I have never flown an AC where you had an option other than cranking the gear down by hand, haha.

Answer 1

[Star Strider]

This seems to work on the test data I’ve given it:

sortedresults = randi(5, 35, 3);                        % Create Data
sortedresults(:,2) = sortedresults(:,2)-1;              % Allow zeros in ‘Land’

DifLand = [diff(sortedresults(:,2)); 0]                 % Calculate changes in ‘LAND’ state
DxLand = [(1:length(sortedresults))' sortedresults(:,2)  DifLand  sortedresults(:,3)]   % Diagnostic check matrix
Land3ix = find((sortedresults(:,2) == 3) & (ismember(DifLand,[-2 -1 +1])))  % Find indices of changes in ‘LAND’ meeting criteria
Iloc23 = find((sortedresults(:,3) == 2) | (sortedresults(:,3) == 3))    % Find indices of ILOC meeting criteria
LandIlocTrue = intersect(Land3ix, Iloc23)               % Indices of ‘LAND’ and ‘ILOC’ meeting criteria
Count = length(LandIlocTrue)                            % Count occurrences

I left the trailing semicolons off, so it will display the results of each step. Note that it deals with the indices of the rows of the data. All the functions are core MATLAB and do not require any Toolboxes.

The only line that is not necessary in your application code is the ‘DxLand’ matrix. The first column is the row index, the second is ‘LAND’, the third are the end-zero-padded differences in ‘LAND’ (zero-padded to make the vector equal to the others), and the fourth is ‘ILOC’. I created ‘DxLand’ so I could check to be sure the code worked. I ran the code several times to be sure it worked as I understand your criteria.

[steve]

thanks very much - will give this a try this afternoon.

[Star Strider]

My pleasure!

[steve]

hello again - I think I am close. can you tell me what the zero in the code below is doing?

DifLand = [diff(sortedresults(:,2)); 0]

Thanks!

[Star Strider]

It makes my ‘DifLand’ variable equal in length to the rest so that MATLAB will be happy concatenating it with the other vectors in ‘DxLand’, and using it in the other statements that need it. (The output of the diff function is shorter than its array argument by whatever number of elements the second argument of diff is, so to get the indices to work out correctly, it’s necessary to pad it. Where to pad it depends on the requirements of the application using it. This is an illustration of the ‘art’ of programming.)

[steve]

thanks! Just wanted to understand the purpose of it and that fills the bill!

[Star Strider]

My pleasure!

And be sure you always have ‘three in the green’ on short final!

Answer 2

[Image Analyst]

How about this:

sortedresults = randi(5, 300, 3); % Create sample data.
% Extract columns 2 and 3.
land = sortedresults(:, 2)
ilok = sortedresults(:, 3)
% Find where ilok column has a value of 2 or 3.
ilok2or3 = ilok == 2 | ilok == 3
% Find the change from one element to the next in column 2
diffland = diff(land)
% Find where the value is 3 in land, and the difference is not 0 or -3
% and column 3 = 2 or 3.
land3 = (land(1:end-1) == 3) & (diffland ~=0 & diffland ~= 03) & ilok2or3(1:end-1)
% Sum up the result.
result = sum(land3)

[steve]

thanks, this works. I did not notice before that 2 different folks had responded. Thanks again!

[Image Analyst]

You're welcome. It's good to scroll all the way down the screen to make sure you see all responses.