Extract maximum values from three different vectors when their associated location vectors crosses

Dear all,

I have six vectors, three with peak values and three vectors with associated location values based on three peak analysis.

% Sensor 1 (15 events)
loc1=[1213  1373  1795  2659  4348  5020  5491  5667  8537  8953  9173  10536  10826  10991  11489];
pks1=[404  998  891  1844  329  428  1025  335  2354  549  997  438  7531  614  545];
% Sensor 2 (14 events)
loc2=[1373  1795  2659  3051  4348  5020  5491  5667  8537  8953  9173  10826  10991  11489];
pks2=[1372  1065  1327  323  568  461  989  504  1322  532  440  7650  566  484];
% Sensor 3 (11 events)
loc3=[1373  1795  2659  4348  5491  8537  8953  9173  10826  10991  11489];
pks3=[1547  920  931  601  1058  1628  652  638  5332  741  397];

The following algorithm applies:

• First, select all of the peaks defined by pks1, pks2 and pks3.
• Next, identify the maximum value from pks1,pks2 and pks3 for each "event". The events are given by the location vector in time and shall be defined as "same event" when the time cross +-2 seconds.
• The new vector shall contain all the maximum values from all the identified events. 16 events have been identified by comparing the values in the 3 loc-vectors.
• There may be 1, 2 or 3 peak values in one event.

Example: Only one peak value in an event. I.e. new vector must include 323.

Sensor 2: loc2(4),pks2(4) / time=3051s and peak=323

Example: Two peak values in an event. I.e. new vector must include 1547.

Sensor 2: loc2(1),pks2(1) / time=1373s and peak=1372

Sensor 3: loc3(1),pks3(1) / time=1373s and peak=1547

Example: Three peak values in an event. I.e. new vector must include 1065.

Sensor 1: loc1(3),pks1(3) / time=1795s and peak=891

Sensor 2: loc2(2),pks2(2) / time=1795s and peak=1065

Sensor 3: loc3(2),pks3(2) / time=1795s and peak=920

The output of the new vector shall be:

new_vector=[404 1547 1065 1844 323 601 461 1058 504 2354 652 997 438 7650 741 545]

I have attached a plot illustrating the identified events from the different sensors.

If anything is unclear please let me know!

Thanks in advance.

[darova]

Did you try something? Can you show?

[Askeladden2]

Darova,

This problem arised for a couple of days ago, but every attempt I have tried has failed so it is no point of presenting anything.

I initially had problems with only two variables (two sensors), but the problem extended to three variables when the tests had been performed. I received help from a very friendly guy (@Adam Kunz) on the first matter.

[Askeladden2]

Darova,

If I concatenate all the vector values, I'll end up with two vectors, one with all events and the other with peaks.

A, sorted events.

A=[1213 1373 1373 1373 1795 1795 1795 2659 2659 2659 3051 4348 4348 4348 5020 5020 5491 5491 5491 5667 5667 8537 8537 8537 8953 8953 8953 9173 9173 9173 10536 10826 10826 10826 10991 10991 10991 11489 11489 11489];

B, associated peaks.

B=[404 1547 1372 998 920 1065 891 1503 1844 1627 323 329 568 601 428 461 1025 1010 1058 335 504 2354 1473 1628 652 556 549 997 440 638 438 5332 7650 7531 741 596 614 545 545 457 ]

Is it possible to use a loop of vector A in order to get the maximum value of vector B, when vector has the same (time+-2s)?

I.e. If A(x).. <= (A(x+1)+-2)

C(x)=max(A(x),A(x+1);

else

C(x)=(A(x);

Example:

For example is A(1,1) unique and therefore C=B(1,1).

For A(2,1) this is equal (+-2s) to A(3,1) and A(4,1). Therefore C=max(B(2,1),B(3,1),B(4,1))

I am not sure if this is the easiest way or even possible..

[darova]

Sure it's possible

C = A;            % preallocate array
for i = 1:length(A)-1
    if abs(A(i) - A(i+1)) < 2
        C(i) = max( A(i), A(i+1) );
    end
end

But another part of your task remains unclear for me

Answer 1

[the cyclist]

% Sensor 1 (15 events)
loc{1}=[1213  1373  1795  2659  4348  5020  5491  5667  8537  8953  9173  10536  10826  10991  11489];
pks{1}=[404  998  891  1844  329  428  1025  335  2354  549  997  438  7531  614  545];
% Sensor 2 (14 events)
loc{2}=[1373  1795  2659  3051  4348  5020  5491  5667  8537  8953  9173  10826  10991  11489];
pks{2}=[1372  1065  1327  323  568  461  989  504  1322  532  440  7650  566  484];
% Sensor 3 (11 events)
loc{3}=[1373  1795  2659  4348  5491  8537  8953  9173  10826  10991  11489];
pks{3}=[1547  920  931  601  1058  1628  652  638  5332  741  397];

all_loc = [loc{1} loc{2} loc{3}];
all_pks = [pks{1} pks{2} pks{3}];

num_loc = numel(all_loc);

single_loc = [];
single_pks = [];
for ni = 1: num_loc
   [isCloseLoc, idxClose] = ismembertol(all_loc(ni),single_loc,2,"DataScale",1);
   if isCloseLoc
       single_pks(idxClose) = max(single_pks(idxClose),all_pks(ni));
   else
       single_loc = [single_loc, all_loc(ni)];
       single_pks = [single_pks, all_pks(ni)];
   end
end

figure
hold on
for nc = 1:3
    plot(loc{nc},pks{nc},'.')
end


figure
hold on
for nc = 1:3
    plot(single_loc,single_pks,'.')
end

Note that I used a cell array instead of dynamically named variables.

[Askeladden2]

Thank you thecyclist!

I am extremly grateful!