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

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 !!!

Answer 1

[dpb]

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., ...

[Sud Mudiyanselage]

yes correct, my mistake there on second exmaple answer. Thanks Thanks heaps dpb :)

Can you explain definition N and ==2 meant to do ? does it only check starting from first # neg to pos sign or both pos to neg too ? And, max(A(ix)) is looking at max of col 1 not col 2, yes?

example 3 , all are negative (no sign change) , take max of col 2 (2.333), find col 1 (-0.23) - 3 rows back.

A=[-5.67 1.999

-5.66 0.994

-0.15 2.024

-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];

answer should be -3.00 1.425

example 4 , if sign change from positive to negative , at negative values of col 1 (-5.19 and -0.23) then look of column 2 ((1.065 and 2.333), find the max (2.333) between those values found in col 2, and find matching value column 1 (-0.23) and go 3 rows back.

A=[5.77 -1.999

5.66 -0.994

0.25 -2.024

5.38 -1.049

-5.19 1.065

-4.42 1.175

-5.02 1.475

1.81 -1.393

0.68 -1.342

-0.23 2.333

-0.44 2.166];

answer should be : -5.02 1.475

In the real problem, I have 4 conditional tests (exmaple 1, example 2, example 3 and example 4) satisfy only one set of data, and find one answer. If I want to combined all condiotnal tests into one loop and test on a differnt set of data which will satisfy one of them. there is an error in condition 1 and 4. if yu can help... thanks in advance again !!!

if N==1 % no sign change occurances  
    ix=find(diff(sign(A(:,1)))==0,N)+1;
    A(ix+7,:) % code condition 1 not working 
if N==1 % sign change occurance only once, sign change =negative to positive
    ix=find(diff(sign(A(:,1)))==2,N)+1;
    A(ix-3,:)
if N>=1 % sign change occurance more than once, sign change negative to positive 
    ix=find(diff(sign(A(:,1)))==2,N)+1;
    [mx,imx]=max(A(ix));
    A(ix(imx)-3,:)
elseif N>=1 % sign change occurance more than once, sign change positive to positive 
    ix=find(diff(sign(A(:,1)))==2,N)+1;
    [mx,imx]=max(A(ix)); 
    A(ix(imx)-3,:) % code condition 4 not working
    
end   

[dpb]

The first case result would be that the result of the call to find will be empty -- there are/will be no cases that will satisfy the logical test. BUT, then it also depends upon what is the definition of "sign change" looking for -- the original Q? definition was from -ive to +ive which is what "==2" finds; "==0" would be all elements are either +/- --but there could also be zero to signed either way and a single transition from +ive to -ive that aren't all elements of the same sign but don't meet the original definition.

But, if all are +ive or -ive, then find(x==0) will return the vector [1:numel(x)-1] since the difference will all be identically zero IF there are no zeros; sign returns 0 for 0, not 1; the MATLAB implementation considers 0 as being unsigned. You'll also have to deal with the edge case if it can/does exist.

There is a typo; I did make the error initially that I didn't check for the value in the third (second in example since I just kept the double data for demo purposes) column -- I thought I had fixed that but obviously copied over the wrong line and didn't notice. Put the column in the indexing expression

[mx,imx]=max(A(ix,3));

The above logic will NOT find the location of the second -ive to +ive transition; that wasn't in the original problem definition. But, there can't be a sign change to +ive without a -ive so the index vector ix will return as many of those as are in the input column up to the number requested in N. So, if there were two or more in the overall column, picking the first two would still give that particular set of two and the location of the latter would be the second index in ix. But, what would it then mean about the maximum -- that isn't so obvious to me what you would expect.

N is used in the original answer to separate between the two cases asked for as an input condition for the search where the condition is one known a priori; if the point is instead to characterize the series into one of the above categories, then the test for each possible result would need to be done -- and the fourth is not mutually exclusive against the third; it cannot happen unless the other is also true so who wins?

The above logic to return the first three asked for locations relative to the sign changes can be generalized with the N=2 logic by

1. ix is empty --> no changes
2. 1 -ive to +ive change --> returns value requested
3. 2 or move -ive to +ive change --> returns value requested relative to the second location

That answers the Q? as posed; if it is a question instead regarding characterizing the whole vector, then remove the N count limitation on the nunber of values possibly returned by find and count the numel(ix) returned and decide what to do based on that; it can be from 0 (empty) to size(A,1)/2 if the input vector were alternating -+ every other element).

[Sud Mudiyanselage]

Thanks for your reply dpb, i am trying to absorb what you have said.

did you say, if the data at Col 1 chnage from positive to negative ( example 4), code provided at example 2 would not work, because it is set up to follow sign chnage neg --> pos not pos --> neg ..correct ?

I am still struggling a abit to put together a loop and that wuill all 4 conditional tests for a any given data set.

[dpb]

Yes, that was the original Q? asked, so that was the problem solved..just try and see with a simple test case or two...

x=[5.2 -4.8];
sign(x)

ans = 1×2

1 -1

diff(sign(x))

ans = -2

What's agonna' happen if it is

x=[-3.2 8.3];

instead?

[dpb]

" put together a loop and that wuill all 4 conditional tests for a any given data set"

But there's a problem in definition of what those conditions are in that the four cases as you laid them out aren't orthogonal.

You first have to define a set of conditions that are able to be uniquely separated that one dataset can't match more than one thing.

See the earlier comment again that outlines at least some of the problems that aren't defined clearly what the intent is nor the result for those cases should be.

[Sud Mudiyanselage]

apppologies for the late reply dpb. thanks for your reply.

my intention was to create four random data set exmaples here ( independent from each other) to test each condition independently, they are not meant to give same answer. ( but i gathered two examples give the same answer, which was my mistake) as long as the data set satifies only one of the four conditions.

So, for the real problem i will have only one set of data, that will satisfy only one from the 4 conditions, the reason for the for loop.

[dpb]

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]

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'));