Detecting a Continuing Increase/Decrease, or Stabilizing Trend in Data

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Jumping into the question: so I have some data in somewhat like this:
A = [1 2 1 2 1 2 3 4 5 6 7];
basically it will start to increase at some point, and I'd like to know where that 'some point' is. For the example above, is there a code/function that lets MatLab detect the increase in consecutively 3 numbers (6th element) and return the position?
The final purpose is to delete the parts before the consecutive increase and keep the parts after it; for now I'd really like to understand that part;
Thanks in advance,
  1 Comment
Image Analyst
Image Analyst on 28 Jun 2016
The signal starts increasing, with a run of 3 or more, at element #5, with the number 1. Why did you say it starts at element number 6?

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Answers (3)

Image Analyst
Image Analyst on 28 Jun 2016
This code is fairly robust. It finds all runs where it's monotonically increasing for 3 elements or more and prints out information about that run. It can handle more than one run, or no runs at all, and gets info on all existing runs.
A = [1 2 1 2 1 2 3 4 5 6 7 5 8 0 1 3 6 9 4];
da = diff(A)>0
stats = regionprops(bwlabel(da), 'Area', 'PixelIdxList')
% Find which runs have a length of at least
% 3 elements of monotonically increasing elements
threeOrLonger = find([stats.Area] >= 3)
% Print out those sequences.
if ~isempty(threeOrLonger)
for blobIndex = 1 : length(threeOrLonger)
% This blob is 3. Find where the blob starts and stops.
startingIndex = stats(threeOrLonger(blobIndex)).PixelIdxList(1);
endingIndex = stats(threeOrLonger(blobIndex)).PixelIdxList(end)+1;
% Print out that run.
fprintf('For run #%d, startingIndex = %d, endingIndex = %d, the elements = ', ...
blobIndex, startingIndex, endingIndex);
fprintf('%d', A(startingIndex : endingIndex));
msgbox('No run is 3 or longer');

Superficial on 12 Mar 2020
I know I'm late to this thread, but this is my solution to this problem. Perhaps not as elegant but easy to understand.
I wanted to determine the value (in y) and timepoint (in x) of the start of a reaction.
for i=2:length(y)-1 % Exclude first and last values in series so y(i-1) and y(i+1) actually do exist
if y(i-1)<y(i) && y(i)<y(i+1) % finds 3 consecutive rising values
end % Ends the 'if' statement at this point.
% i is therefore the index of the (middle of 3) rising values in y
% In my case I wanted the x value as well to define the x (time) value at the same point

KSSV on 28 Jun 2016
You can calculate diff of diff of A, when it remains constant, form that position the function increases. Eg:
A = [1 2 1 2 1 2 3 4 5 6 7];
K = diff(diff(A)) ;
idx = find(K==0) ;
iwant = idx(1)+1
Also go through monotonically increasing functions and monotonically decreasing functions.
x = [1 2 3 4 5 6]
all(diff(x)>0) % increasing
% But now:
x(7) = 3


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