Does a way exists to get the results of the retime function in matlab 2015b??? Any help would be great!!!

I'm trying to have one monotonic time created taking the end of the 1st cell of time data and continuing on until the last cell time data as one continous time. the frequency = 0.055;

(e.g. last cell data of cameraTime{1,1} = 3.2969e+03; cameraTime{1,2} = 3.2969e+03 + 0.055 and continue with the size of time data in the remaining cells to create one time for all the total cells.)

Answer 1

[Walter Roberson]

create cell arrays in which the camera time is the first column and the data corresponding to that time is in the other columns. cell2mat() those together into one big matrix and sort() on time.

Create a vector of times from the first to ceil((last-first)/0.055)*0.055+first at intervals of 0.055

Now do interp1() over the arrays, original time values as the x values, vector with consistent offset as the x values to query at.

[Brandon Leonard]

Thank you Walter for the quick response. I'm fairly new to matlab so I feel I left out valuable information on my question.

#1 the time is collected in "CAN time" and has a range from 0 to 4400 where the times of the cells can begin new once the maximum time is reached which I provided a visual with the attached images.

#2 The frequency is not a finite number of 0.055 but that is more of an average frequency which I don't know how big of a deal this would be.

I understand the creating of the cell arrays with the camera time and the data(camera) and using the function cell2mat() to in one matrix.

My questions which I ask that you please explain further in detail?

#1 Sort() on time meaning use this syntax [B,I] = sort(___) I assume??

#2 Could the creating of times from first to ceil still be used because of the starting over of time? Also could you give a visual of the line ( first to ceil((last-first)/0.055)*0.055+first at intervals of 0.055).

#3 Would the frequency not exactly being a finite number of 0.055 change the above equation?

Any help would be awesome!!!

[Walter Roberson]

When the times "start over", then should all of the entries with the same time marker be considered to be the same time? For example, if two entries had time 2.785 but one was in cell 1 and the other was in cell 3, then should those be considered the same time 2.785 ?

Or should it be treated sort of like clock time, where each time that a value is detected that is lower than the immediately preceeding value, that it should be assumed that your readout has wrapped into the next cycle, and that the cycles need to be distinguished? So for example that if it went 4398, 4400, 2, 4, that the later entries should have 4400 added to them, making them 4398, 4400, 4402, 4404 ?

If the frequency is not a definite 0.055 then how do you know what monotonic time interval you want to create? My assumption here was that some of the readings might have been read at slightly the wrong time, or that a couple of readings might be missing, and that the task was to use the existing data to create new data that is at exact intervals of 0.055 . This creation of new data at exact intervals can be done as long as it is know what the interval is. But perhaps you are really just asking about how to handle unwrapping over the 4400/0 boundary so that all of the times are in increasing order?

[Brandon Leonard]

Yes, your second statement is what I'm trying to do treating if like clock time(Or should it be treated sort of like clock time, where each time that a value is detected that is lower than the immediately preceeding value, that it should be assumed that your readout has wrapped into the next cycle, and that the cycles need to be distinguished? So for example that if it went 4398, 4400, 2, 4, that the later entries should have 4400 added to them, making them 4398, 4400, 4402, 4404 ?) because the data is being collected and the corresponding time where each day is stored into a folder. My overall goal is to study the behavior in the change of data over a long period of time (e.g. one month) but I need the time to be one time line. It can begin with cameraTime{1,1}....cameraTime{1,n} ending with the end of the last data in the folder. Hopefully this hepls you.

[Walter Roberson]

Is each cameraTime a different cycle (e.g., a different day) with at most one cycle per entry? Or are there cycle transitions in the middle of cells?

[Brandon Leonard]

The data being collected is the output of a camera(so for every data output is a corresponding time output of equal sizes (e.g. camera{1,1} cameraTime{1,1}). The total of the all cameraTime is one day where each cameraTime is the duration of one period of data collected during the day.

[Walter Roberson]

put them all into a single matrix. Then add to the times

duration_of_cycle * cumsum([0;diff(Times_column)<0]);

[Brandon Leonard]

#1 By "duration_of_cycle" do you mean using the function duration on the time?

#2 When you say "Then add to the times" do you mean add as displayed below?

for i = 1:segment
 time = cell2mat(cameraTime(i)); 
  end
D = duration(time,'Format', 'hh:mm:ss.SSS');
final time = time + (D * cumsum([0;diff(time)<0]);

[Walter Roberson]

duration of cycle is 4400. You could use duration objects in this matter, or you could just leave everything numeric. As in

4399.989
4399.9945
   0.0
   0.0055

going to

4499.989
4399.9945
4400.0
4400.0055

[Brandon Leonard]

Thanks Walter for your time. I didn't exactly use your answer but it put me on a path to find a suitable solution. By taking the difference of times (cam{i}) I was able to add the iterations by adding the end number to dt because the first position in column is zero. I will gladly accept your answer because you are a great help!!

dX1 = 0; dt = 0.055

for i = 1:segment

            cam{i} = (cameraTime{i}-cameraTime{i}(1));
            cam{i} = cam{i} + dX1;
            camRT{i} = cam{i};
           dX1 = cam{i}(end,1) + dt;