Check if number of changes from negative to positive on data table and perform set of conditional tests

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Conditional testing example 1 as follows:
18:41:40 -5.67 0.999
18:41:40 -5.66 0.994
18:41:40 -5.55 1.024
18:41:40 -5.38 1.049
18:41:40 -5.19 1.065
18:41:40 -4.42 1.175
18:41:40 -3.00 1.425
18:41:40 -1.81 1.393
18:41:40 0.68 1.342
18:41:40 0.23 2.333
18:41:41 0.44 2.166
if number of changes @ column 2 =1 , at the row # of change go 3 rows back of timestamp on column 1 and get the matching column 2 value. (answer should give -4.42)
Conditional testing example 2 as follows:
19:41:40 -5.67 0.999
19:41:40 -5.66 0.994
19:41:40 0.15 1.024
19:41:40 -5.38 1.049
19:41:40 -5.19 1.065
19:41:40 -4.42 1.175
19:41:40 -3.00 1.425
19:41:40 -1.81 1.393
19:41:40 0.68 1.342
19:41:40 0.23 2.333
19:41:41 0.44 2.166
If number of changes@ column 2 >1 and , then look at each change where gives highest value from column 3 and get the the corresponding timestamp in column 3 and go 3 rows back and get corresponding column 2 value. (answer should give -5.19)
Table doesn't have column headings .
please help, Thank you !!!

Accepted Answer

dpb
dpb on 1 Oct 2022
Edited: dpb on 2 Oct 2022
A=[-5.67 0.999
-5.66 0.994
-5.55 1.024
-5.38 1.049
-5.19 1.065
-4.42 1.175
-3.00 1.425
-1.81 1.393
0.68 1.342
0.23 2.333
0.44 2.166];
% the engine
% case 1
N=1;
ix=find(diff(sign(A(:,1)))==2,N)+1;
A(ix-3,:)
ans = 1×2
-4.4200 1.1750
A=[-5.67 0.999
-5.66 0.994
0.15 1.024
-5.38 1.049
-5.19 1.065
-4.42 1.175
-3.00 1.425
-1.81 1.393
0.68 1.342
0.23 2.333
0.44 2.166];
% case 2
N=2;
ix=find(diff(sign(A(:,1)))==2,N)+1;
[mx,imx]=max(A(ix,3));
A(ix(imx)-3,:)
ans = 1×2
-4.4200 1.1750
NB: I believe either the first or the second answer is wrong; probably the second. -5.19 is four places preceding the location of the 2nd positive in the second case. The two results given are inconsistent with each other in the offset counting from the location of the positive value after the change...
Above will have to have error checking, etc., if the result of the find() is empty, etc., etc., ...
  8 Comments
dpb
dpb on 8 Oct 2022
Edited: dpb on 8 Oct 2022
But the problem is the conditions themselves aren't unique so a data set that satisfies one also satisfies another -- who wins? Unless some other criterion is added to differentiate those, whichever is first in the test sequence will be the winner. Left in that condition (so to speak :) ), then there's no point in having the other condition; it'll never be satisfied so might as well not be there at all.
You've got the ideas of how to make such tests for the condition of sign changes; you've just got to decide what it is that you actually want the results to be and structure the code/refine the condition definitions to provide those results, but at this point you've not defined what is the actual result to be returned in the multiple sign changes case vis a vis just one for at least one conundrum to be resolved.
Sud Mudiyanselage
Sud Mudiyanselage on 17 Oct 2022
Edited: Sud Mudiyanselage on 17 Oct 2022
hi dpb ... Thanks for your reply.
Yes i understand what your saying. I actually managed to fix the sign change from positive to negative. it is now working ....SO THANK YOU for your code helped me to get that last bit working .....thx heaps !!!
I do have another another issue for a different data set, hopefully you can give some guidance. i have a set of data attached (.txt file) represents heights vs time.
I need to check within the whole profile, how many times the 25, 70 and 99 meter height peak occurs ( i am not after # of occurences). Answers should be as follows as shown in the image:
1. 25 meter height loading - 4 times (red color)
2. 70 meter height loading - 2 times (green color)
3. 99 meter height loading - 1 time ( pink color)
I have tried using following methods but neither had luck, maybe i need to do some looping to check .... i am not sure. thanks in advance.
[pks_25,locs_25]=findpeaks(data,'MinPeakheight',25);
Num_peaks_25=(findpeaks(data,'MinPeakheight',25,'MinPeakDistance',n));
sum(islocalmax(data,'FlatSelection','First'));

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