How to compare two vectors that don't have equal length?

Based on the first few elements that you mentioned, probably the first option is what you want. No matter what, you're going to have one element of x2 that's left out.

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Voss 2021-12-13

在 MATLAB Online 中打开

In that case, a good approach would be to take the difference of the index of each peak found by one method and the index every peak found by the other method. The difference may not be exactly zero for the peak to be the 'same' one found by both methods, but we can look at how far each peak in the data set with 535 peaks is from its nearest peak in the data set with 500 peaks. Then if they are close, they are the 'same', and if they are too far apart they are different, indicating that the method that found 500 missed that one.

Here is some code that does that, with comments for explanation:

load('IndexManualx17.mat'); % 500 peaks
load('PT_index_raw_x17.mat'); % 535 peaks
% get a 535-by-500 matrix of the distance from each peak found by one 
% method to every peak found by the other method (distance in terms of 
% absolute difference of the indexes):
didx = abs(PeakIndex_PT_x17-PeakIndexx17.');
% get the minimum distance each peak in the 535 set is away from a peak in
% the 500 set:
d_min = min(didx,[],2);
% any peak that is closer than a certain amount to its nearest peak in the
% other set is considered to be the same peak, and any peak further than a
% certain amount from its nearest peak in the other set is considered to be
% different, but we don't know what those thresholds might be, so:
% plot a histogram of this minimum distance:
figure();
hist(d_min,100);
% the histogram shows most peaks are between 4 and 7 sample indexes away
% from their nearest peak in the 500 set, but a handful are around 35-40
% samples away. this is a pretty clear distinction, so:
% find the peaks in the 535 set that are more than 20 samples away from 
% their nearest peak in the 500 set (could use any threshold between 7 and 
% 34 for these particular data):
idx = find(d_min > 20);
% these are the 35 peaks that the 500 set missed, and they occur at
% PeakIndex_PT_x17(idx) sample numbers
% visualization:
figure();
% plot the 535 indexes where peaks were found in the 535 set (with blue circles):
plot(PeakIndex_PT_x17,ones(1,numel(PeakIndex_PT_x17)),'bo');
hold on
% plot the 500 indexes where peaks were found in the 500 set (with black x's):
plot(PeakIndexx17,ones(1,numel(PeakIndexx17)),'xk');
% plot the 35 from the 535 that aren't in the 500 (with red circles):
plot(PeakIndex_PT_x17(idx),ones(1,numel(idx)),'ro');
% now zooming way in should reveal that most of the blue circles are very 
% close to a black x and occasionally there is a red circle with no 
% corresponding black x:
set(gca(),'XLim',[62000 64000]);
% pan around to see more
pan();

Malahat Mehraban 2021-12-13

Thank you Benjamin, it works great.

What is your opinion about using the time of peak occurred instead of using an index? actually, I want to use a 128-millisecond tolerance for the comparison. If the detected peak is 128 milliseconds sooner or later than the detected peak of the other method, it will determine as a wrong detection.

Voss 2021-12-14

Assuming those data are regularly-sampled, it shouldn't matter whether you use index or time.

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How to compare two vectors that don't have equal length?

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How to compare two vectors that don't have equal length?

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