Interp1 error: Can someone please help.

I have a time series matrix that is not evenly spaced in time.

I have used the following commands to re-sample it with constant time increment (60sec increment). This command works. I have checked the datevector and the time values are incrementing every minute

New 1min sampled time series

downsampled_T= datenum(y0,m0,d0,h0,min0,(s0:60:s0+du))';_

Original timeseries in datenum

oldT=datenum(T);

Interpolation of original values

sparseX=interp1(oldT,D1_position(:,7),downsampled_T);

I get the following error: Error using griddedInterpolant The grid vectors are not strictly monotonic increasing.

Any ideas why? My datevec seems to be incrementing evenly. But my datenum seems to be not?

Answer 1

[Wayne King]

With the precision that you have it seems that your oldT vector is not monotonic.

For example:

   X = 1:9;
   X = [X 9];
   Xnew = 1:0.01:9;
   y = randn(9,1);
   ynew = interp1(X,y,Xnew);

gives that error because X(end) and X(end-1) are the same value.

[Bedanta]

how do I rectify this issue with precision?

I seem to have expected values for the datevec.

[Bedanta]

Another doubt is that, the same command works on a previous version of my raw matrix where I had few more rows. But I am unable to narrow it down to what is different in the new matrix that causes the commands to fail

Answer 2

[Markus Schmidt]

I had a similar problem. Handling with a big dataset (where x is the time) I experienced same behaviour. I figured out a solution. Even though I applied 'unique' to my data matrix, some of the entries are 'not unique' regarding subtracting them (which is used for linear interpolation I guess.) So I applied the following algorithm to my data:

% Create matrix with dataset of properties to be interpolated (northing,
% easting, heading, velocity)
% ts_interp = test_series(isnan(test_series(:,2)),:);
ts_interp = unique(test_series(isnan(test_series(:,2)),[1 9 10 11 12]),'rows');

% For interplation, the function unique is not sufficient to delete enough
% entries. The time data will be checked for differences of 0, which
% depends on the machine precision.
delta = diff(ts_interp(:,1));
nodiff = find(delta == 0);

for k=1:length(nodiff)
    ts_interp(nodiff(k),1) = NaN;
end

ts_interp = ts_interp(~isnan(ts_interp(:,1)),:);