Attempting to find data points of two different matrices based off of time, but outcome is that I am getting the same value repeatedly

I have one matrix with data in the first column and time stamps in the second column (datamatrix.mat). The next matrix contains spiketimes (spiketimematrix.mat). I want to get the data point in the first column of first matrix that is the closest time point corresponding to the spike times in spiketimematrix.mat. For example, the first spiketime is 166.1670, which corresponds to the closet time point of 166.1696 and corresponds with the data point 2.5281. I have tried using the following code:

for k=1:length(spiketimesmatrix)
  dataout(k)=datamatrix(find(spiketimesmatrix(k)>=datamatrix(:,2),1));
end

However, I am getting the first value of =1.8577 repeatedly and I am not sure why?

Answer 1

[dpb]

'Cuz you're not restricting the search to only begin with the last found location and by default find will find the first location that satisfies the condition. If the data are ordered in ascending order, then once you have a location you're looking for that is greater, then any other that is that larger or greater will also satisfy...

Two ways to fix--

1. Use the 'last' optional argument and search for '<=' instead of '>=' which will give you the last value that isn't over, or
2. Use the search for a minimum difference instead, which actually is what your topic says you're looking for, anyway. Note the search for the first/last over/under a given value isn't necessarily the closest, only the one before the sign change of the difference.

for k=1:length(spiketimesmatrix)
  dataout(k)=datamatrix(find(min(abs(spiketimesmatrix(k)-datamatrix(:,2))),1));
end

[Krispy Scripts]

My elements in spikematrix and datamatrix are not equal, so I a getting the an error for this. Do my two matrices have to be the same size for this to work?

[dpb]

No, but got to make sure you don't try to index using the longer into the shorter...here's a working sample written without find...

>> spike=linspace(0,100,13).';    % a sample list of times
>> data(:,2)=linspace(1,155,6).'; % different length data array
>> data(:,1)=1:length(data);      % just an id to look at
>> spike
spike =
       0
  8.3333
 16.6667
 25.0000
 33.3333
 41.6667
 50.0000
 58.3333
 66.6667
 75.0000
 83.3333
 91.6667
100.0000
>> data
data =
  1.0000    1.0000
 31.8000   31.8000
 62.6000   62.6000
 93.4000   93.4000
124.2000  124.2000
155.0000  155.0000
>> for i=1:length(spike)
     [~,ix]=min(abs(spike(i)-data(:,2)));  % find nearest location
     disp([spike(i) ix data(ix,:)])  % show results
  end
   0     1     1     1
  8.3333    1.0000    1.0000    1.0000
 16.6667    2.0000    2.0000   31.8000
 25.0000    2.0000    2.0000   31.8000
 33.3333    2.0000    2.0000   31.8000
 41.6667    2.0000    2.0000   31.8000
 50.0000    3.0000    3.0000   62.6000
 58.3333    3.0000    3.0000   62.6000
 66.6667    3.0000    3.0000   62.6000
 75.0000    3.0000    3.0000   62.6000
 83.3333    4.0000    4.0000   93.4000
 91.6667    4.0000    4.0000   93.4000
100.0000    4.0000    4.0000   93.4000
>>

Hopefully that mimics your situation well enough you can adapt it...note I made the granularity purposefully large to illustrate the min() selection of which point chosen...

[Krispy Scripts]

Great! Can I output this into a matrix instead of displaying it?

[dpb]

Sure, why not? Simply use a LHS variable; good practice will preallocate based on size(spike) array by number of columns needed.