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I got a good answer to my question:

Here is my MATLAB version:

MATLAB Version 9.8 (R2020a)

Communications Toolbox Version 7.3 (R2020a)

DSP System Toolbox Version 9.10 (R2020a)

Instrument Control Toolbox Version 4.2 (R2020a)

LTE Toolbox Version 3.3 (R2020a)

MATLAB Compiler Version 8.0 (R2020a)

Parallel Computing Toolbox Version 7.2 (R2020a)

RF Toolbox Version 3.8 (R2020a)

Signal Processing Toolbox Version 8.4 (R2020a)

Symbolic Math Toolbox Version 8.5 (R2020a)

But I neglected to say that I use dY_idx and Y_adj_close in a subsequent function call where both arguments must have the same dimensions.

Here is my code that keeps the dimensions the same. Is there any way to improve this code to make it run faster (for large vectors)?

Y = [ 1.1 1.12 9.2 8.3 8.295 8.292 4.1 4.12 4.19];

plot(Y), hold on;

% Original 2 lines of code that loses the right-most point is:

% dY_idx = find(abs(diff(Y)) < 0.022);

% Y_adj_close = Y(dY_idx)

% Here are my 5 lines of code that saves the right-most point to keep the right-most point of each grouping.

% Can these 5 lines be reduced?

logidx = abs(diff(Y)) < 0.022;

xLog = [logidx 0] | [0 logidx];

Y_adj_close = Y( xLog );

dY_idx = 1:length(Y);

dY_idx = dY_idx( xLog );

plot(dY_idx, Y_adj_close, 'r.'), hold off

Paul Hoffrichter
on 8 Jun 2020

Tommy
on 10 Jun 2020

logidx = abs(diff(Y)) < 0.022;

dY_idx = find([logidx 0] | [0 logidx]);

Y_adj_close = Y( dY_idx );

Seems to me that every part of these three lines is required, but maybe it can be made even faster.

Of course number of lines alone isn't the only thing to consider. I'd imagine this would be slower:

dY_idx = find([abs(diff(Y)) < 0.022 0] | [0 abs(diff(Y)) < 0.022]);

Y_adj_close = Y( dY_idx );

KSSV
on 8 Jun 2020

Paul Hoffrichter
on 9 Jun 2020

Just to be clear, I am only interested in whether two adjacent points are within a tolerance. For example, if I have 3 adjacent data points, y1, y2, y3 with following data:

y2 is close enough to y1

y3 is not close enough to y2

y3 is close enough to y1

then outcome should be to select {y1, y2} and corresponding indices. Even though y3 is close enough to y1, it is not adjacent so it is disqualified. My recollection of kd tree alg is that all three points would be accepted, but I could be wrong. And yes, I suspect that for this 1D case, the kd tree or other general algorithms would be overkill. Based on the nature of the data, I do not want y3 to be included.

Here is the big question: Can the OP code be reduced from 5 lines lines of code to 4 or less? The test for success is whether the resultant plot is identical to the OP plot. Often, less lines of code without using for-loops are better optimized by MATLAB, but I could be wrong on that account since I am fairly new to MATLAB.

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